(i)     Write down the value of n .

(ii)    Find the value of m , of p , and of q .

Find \({\rm{P}}(B')\) .

(i) \(n = 0.1\)     A1     N1

(ii) \(m = 0.2\) , \(p = 0.3\) , \(q = 0.4\)     A1A1A1     N3

[4 marks]

appropriate approach

e.g. \({\rm{P}}(B') = 1 - P(B)\) , \(m + q\) , \(1 - (n + p)\)     (M1)

\({\rm{P}}(B') = 0.6\)     A1     N2

[2 marks]

Most candidates were able to find the correct values for the Venn diagram. Unfortunately, however, there were many candidates who did not understand what each region of the diagram represents. For example, a very common error was thinking that \({\rm{P}}(B) = p\) , rather than the correct \({\rm{P}}(B) = p + n\) . 

Candidates seemed to understand the idea of the complement in part (b), but some were not able to find the correct answer because of confusion over the separation of the different regions in the diagram.

Write down the probability that the first marble Anna selects is red.

Find the probability that Anna selects two red marbles.

Find the probability that one marble is red and one marble is blue.

Note: In this question, method marks may be awarded for selecting without replacement, as noted in the examples.

\({\rm{P}}(R) = \frac{6}{8}\left( { = \frac{3}{4}} \right)\)     A1     N1

[1 mark]

attempt to find \({\rm{P(Red)}} \times {\rm{P(Red)}}\)     (M1)

e.g. \({\rm{P(}}R{\rm{)}} \times {\rm{P(}}R{\rm{)}}\) , \(\frac{3}{4} \times \frac{3}{4}\) , \(\frac{6}{8} \times \frac{5}{7}\)

\({\rm{P}}(2R) = \frac{{36}}{{64}}\left( { = \frac{9}{{16}}} \right)\)     A1     N2

[2 marks]

METHOD 1

attempt to find \({\rm{P(Red)}} \times {\rm{P(Blue)}}\)     (M1)

e.g. \({\rm{P(}}R{\rm{)}} \times {\rm{P(}}B{\rm{)}}\) , \(\frac{6}{8} \times \frac{2}{8}\) , \(\frac{6}{8} \times \frac{2}{7}\)

recognizing two ways to get one red, one blue     (M1)

e.g. \({\rm{P}}(RB) + {\rm{P}}(BR)\) , \(2\left( {\frac{{12}}{{64}}} \right)\) , \(\frac{6}{8} \times \frac{2}{7} + \frac{2}{8} \times \frac{6}{7}\)

\({\rm{P}}(1R,1B) = \frac{{24}}{{64}}\left( { = \frac{3}{8}} \right)\)     A1     N2

[3 marks]

METHOD 2

recognizing that \({\rm{P}}(1R,1B)\) is \(1 - {\rm{P}}(2B) - {\rm{P}}(2R)\)     (M1)

attempt to find \({\rm{P}}(2R)\) and \({\rm{P}}(2B)\)     (M1)

e.g. \({\rm{P}}(2R) = \frac{3}{4} \times \frac{3}{4}\) , \(\frac{6}{8} \times \frac{5}{7}\) ; \({\rm{P}}(2B) = \frac{1}{4} \times \frac{1}{4}\) , \(\frac{2}{8} \times \frac{1}{7}\)

\({\rm{P}}(1R,1B) = \frac{{24}}{{64}}\left( { = \frac{3}{8}} \right)\)     A1     N2

[3 marks]

Candidates did very well on parts (a) and (b) of this probability question.

Candidates did very well on parts (a) and (b) of this probability question, and knew to multiply the probabilities of independent events in part (b). However, in part (c), very few candidates considered that there are two ways to draw one red and one blue marble, and therefore did not earn full marks on this question. There were also some candidates who tried to add, rather than multiply, the probabilities in parts (b) and (c).

Candidates did very well on parts (a) and (b) of this probability question, and knew to multiply the probabilities of independent events in part (b). However, in part (c), very few candidates considered that there are two ways to draw one red and one blue marble, and therefore did not earn full marks on this question. There were also some candidates who tried to add, rather than multiply, the probabilities in parts (b) and (c).

Write down the value of

(i)     p ;

(ii)    q ;

(iii)   r.

Find the value of \({\rm{P}}(A|B')\) .

Hence, or otherwise, show that the events A and B are not independent.

(i) \(p = 0.2\)     A1     N1 

(ii) \(q = 0.4\)     A1     N1 

(iii) \(r = 0.1\)     A1     N1 

[3 marks]

\({\rm{P}}(A|B') = \frac{2}{3}\)     A2     N2

Note: Award A1 for an unfinished answer such as \(\frac{{0.2}}{{0.3}}\) . 

[2 marks]

valid reason     R1

e.g. \(\frac{2}{3} \ne 0.5\) ,  \(0.35 \ne 0.3\)

thus, A and B are not independent     AG     N0

[1 mark]

As the definitions of p and q were not clear to candidates, both responses of p \(= 0.2\), q \(= 0.4\) and p \(= 0.5\), q \(= 0.7\) were accepted for full marks. However, finding r eluded many.

Few candidates answered the conditional probability correctly. Many attempted to use the formula in the booklet without considering the complement, and there was little evidence of the Venn diagram being utilized as a helpful aid.

To show the events are not independent, many correctly reasoned that \(0.3 \ne 0.35\) . A handful recognized that \({\rm{P}}(A|B') \ne {\rm{P}}(A)\) is an alternative approach that uses the answer in part (b). Some candidates do not know the difference between independent and mutually exclusive.

Find the values of k such that \(f(x) = 0\) has two equal roots.

Each value of k is equally likely for \( - 5 \le k \le 5\) . Find the probability that \(f(x) = 0\) has no roots.

METHOD 1

evidence of discriminant     (M1)

e.g. \({b^2} - 4ac\) , discriminant = 0

correct substitution into discriminant     A1

e.g. \({k^2} - 4 \times \frac{1}{2} \times 8\) , \({k^2} - 16 = 0\)

\(k = \pm 4\)     A1A1     N3

METHOD 2

recognizing that equal roots means perfect square     (R1)

e.g. attempt to complete the square, \(\frac{1}{2}({x^2} + 2kx + 16)\)

correct working

e.g. \(\frac{1}{2}{(x + k)^2}\) ,  \(\frac{1}{2}{k^2} = 8\)     A1

\(k = \pm 4\)     A1A1     N3

[4 marks]

evidence of appropriate approach     (M1)

e.g. \({b^2} - 4ac < 0\)

correct working for k     A1

e.g. \( - 4 < k < 4\) , \({k^2} < 16\) , list all correct values of k

\(p = \frac{7}{{11}}\)     A2     N3

[4 marks]

A good number of candidates were successful in using the discriminant to find the correct values of k in part (a), however, there were many who tried to use the quadratic formula without recognizing the significance of the discriminant.

Part (b) was very poorly done by nearly all candidates. Common errors included finding the wrong values for k, and not realizing that there were 11 possible values for k.

On the diagram below, shade the region representing \({\rm{P}}(X > 105)\) .

Given that \({\rm{P}}(X < d) = {\rm{P}}(X > 105)\) , find the value of \(d\) .

Given that \({\rm{P}}(X > 105) = 0.16\) (correct to two significant figures), find \({\rm{P}}(d < X < 105)\) .

     A1A1     N2

Note: Award A1 for vertical line to right of mean, A1 for shading to right of their vertical line.

evidence of recognizing symmetry     (M1)

e.g. \(105\) is one standard deviation above the mean so \(d\) is one standard deviation below the mean, shading the corresponding part, \(105 - 100 = 100 - d\)

\(d = 95\)     A1     N2

[2 marks]

evidence of using complement     (M1)

e.g. \(1 - 0.32\) , \(1 - p\)

\({\rm{P}}(d < X < 105) = 0.68\)     A1     N2

[2 marks]

Most candidates did very well on part (a), shading the area under the normal curve.

Not all candidates realised that the problem could be solved by only using the symmetry of the normal distribution curve and the information given. Some of them saw the need to use tables and others just left it blank.

Candidates were only moderately successful on parts (b) and (c), which required understanding of the symmetry of the curve. Many candidates resorted to formulae or tables instead of reasoning through the question.

Not all candidates realised that the problem could be solved by only using the symmetry of the normal distribution curve and the information given. Some of them saw the need to use tables and others just left it blank.

Candidates were only moderately successful on parts (b) and (c), which required understanding of the symmetry of the curve. Many candidates resorted to formulae or tables instead of reasoning through the question.

Find p .

The cumulative frequency diagram is given below.

Use the diagram to estimate

(i)     the 80th percentile;

(ii)    the interquartile range.

evidence of valid approach     (M1)

e.g. \(92 + 52\) , line on graph at \(x = 31\)

\(p = 144\)     A1     N2

[2 marks]

(i) evidence of valid approach     (M1)

e.g. line on graph, \(0.8 \times 160\) , using complement

\( = 29.5\)     A1     N2

(ii) \({Q_1} = 23\) ; \({Q_3} = 29\)      (A1)(A1)

\({\text{IQR}} = 6\) (accept any notation that suggests an interval)     A1     N3

[5 marks]

Part (a) was generally answered correctly, with most candidates showing a good grasp of cumulative frequency from a table.

A surprising number of candidates had difficulty reading values off the cumulative frequency curve. A common incorrect answer for (b)(i) was 29, indicating carelessness with the given scale. Too many candidates gave 40 and 120 for the quartile values.

Find \({\rm{P}}(A)\) .

Find \({\rm{P}}(B|A)\) .

Find \({\rm{P}}(A \cap B)\) .

\({\rm{P}}(A) = \frac{1}{{11}}\)     A1    N1

[1 mark]

\({\rm{P}}(B|A) = \frac{2}{{10}}\)     A2     N2

[2 marks]

recognising that \({\rm{P}}(A \cap B) = {\rm{P}}(A) \times {\rm{P}}(B|A)\)     (M1) 

correct values     (A1)

e.g. \({\rm{P}}(A \cap B) = \frac{1}{{11}} \times \frac{2}{{10}}\)

\({\rm{P}}(A \cap B) = \frac{2}{{110}}\)     A1     N3 

[3 marks]

Most candidates answered part (a) correctly.

Few candidates used the concept of "B given A" to simply "write down" the answer of \(\frac{2}{{10}}\) . Instead, most reached for the formula in the booklet, with which few were successful.

Few also made the connection that part (c) could be answered using both previous answers. Many found \({\rm{P}}(A \cap B)\) correctly even when answering part (b) incorrectly, although some candidates did not decrease the denominator for the second event.

Write down \({\text{P}}(X > 107)\).

Find \({\text{P}}(100 < X < 107)\).

Find \({\text{P}}(93 < X < 107)\).

\({\text{P}}(X > 107) = 0.24\,\,\,\left( { = \frac{6}{{25}},{\text{ }}24\% } \right)\)     A1     N1

[1 mark]

valid approach     (M1)

eg\(\,\,\,\,\,\)\({\text{P}}(X > 100) = 0.5,{\text{ P}}(X > 100) - {\text{P}}(X > 107)\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(0.5 - 0.24,{\text{ }}0.76 - 0.5\)

\({\text{P}}(100 < X < 107) = 0.26\,\,\,\left( { = \frac{{13}}{{50}},{\text{ }}26\% } \right)\)     A1     N2

[3 marks]

valid approach     (M1)

eg\(\,\,\,\,\,\)\(2 \times 0.26,{\text{ }}1 - 2(0.24),{\text{ P}}(93 < X < 100) = {\text{P}}(100 < X < 107)\)

\({\text{P}}(93 < X < 107) = 0.52\,\,\,\left( { = \frac{{13}}{{25}},{\text{ }}52\% } \right)\)     A1 N2

[2 marks]

Write down

(i)     P(B);
(ii)    P(R).

If the blue face lands down, the die is not rolled again. If the red face lands down, the die is rolled once again. This is represented by the following tree diagram, where p, s, t are probabilities.

Find the value of p, of s and of t.

Guiseppi plays a game where he rolls the die. If a blue face lands down, he scores 2 and is finished. If the red face lands down, he scores 1 and rolls one more time. Let X be the total score obtained.

(i)     Show that \({\text{P}}(X = 3) = \frac{3}{{16}}\) .
(ii)    Find \({\text{P}}(X = 2)\) .

(i)     Construct a probability distribution table for X.

(ii)    Calculate the expected value of X.

If the total score is 3, Guiseppi wins \(\$ 10\). If the total score is 2, Guiseppi gets nothing.

Guiseppi plays the game twice. Find the probability that he wins exactly \(\$ 10\).

(i) P(B) \( = \frac{3}{4}\)     A1     N1

(ii) P(R) \( = \frac{1}{4}\)     A1     N1

[2 marks]

\(p = \frac{3}{4}\)     A1     N1

\(s = \frac{1}{4}\), \(t = \frac{3}{4}\)     A1     N1

[2 marks]

(i) \({\text{P}}(X = 3)\)

\( = {\text{P (getting 1 and 2)}} = \frac{1}{4} \times \frac{3}{4}\)     A1

\( = \frac{3}{{16}}\)     AG     N0

(ii) \({\text{P}}(X = 2) = \frac{1}{4} \times \frac{1}{4} + \frac{3}{4}{\text{ }}\left( {{\text{or }}1 - \frac{3}{{16}}} \right)\)     (A1)

\( = \frac{{13}}{{16}}\)      A1     N2

[3 marks]

(i)

     A2      N2

(ii) evidence of using \({\text{E}}(X) = \sum{x{\text{P}}(X = x)} \)     (M1)
\({\text{E}}(X) = 2\left( {\frac{{13}}{{16}}} \right) + 3\left( {\frac{3}{{16}}} \right)\)    (A1)
\( = \frac{{35}}{{16}}{\text{ }}\left( { = 2\frac{3}{{16}}} \right)\)     A1     N2

[5 marks]

win \(\$ 10 \Rightarrow \) scores 3 one time, 2 other time     (M1)

\({\text{P}}(3) \times {\text{P}}(2) = \frac{{13}}{{16}} \times \frac{3}{{16}}\) (seen anywhere)     A1

evidence of recognising there are different ways of winning \(\$ 10\)     (M1)
e.g. \({\text{P}}(3) \times {\text{P}}(2) + {\text{P}}(2) \times {\text{P}}(3)\) , \(2\left( {\frac{{13}}{{16}} \times \frac{3}{{16}}} \right)\) , \(\frac{{36}}{{256}} + \frac{3}{{256}} + \frac{{36}}{{256}} + \frac{3}{{256}}\)
\({\text{P(win }}\$ 10) = \frac{{78}}{{256}}{\text{ }}\left( { = \frac{{39}}{{128}}} \right)\)     A1     N3

[4 marks]

This was the most difficult of the extended response questions for the candidates. Finding s and t correctly in part (b) was difficult, with many confused between writing appropriate probabilities on a single branch compared to at the final end of a multiple branch. Many candidates had no idea what to write for a probability distribution and those who did often had probabilities that did not sum to 1. Candidates who wrote a probability distribution often could correctly compute the expected value. The final part was the most challenging, but some good answers were seen. The most common error was not recognizing that there were two different ways of winning.

This was the most difficult of the extended response questions for the candidates. Finding s and t correctly in part (b) was difficult, with many confused between writing appropriate probabilities on a single branch compared to at the final end of a multiple branch. Many candidates had no idea what to write for a probability distribution and those who did often had probabilities that did not sum to 1. Candidates who wrote a probability distribution often could correctly compute the expected value. The final part was the most challenging, but some good answers were seen. The most common error was not recognizing that there were two different ways of winning.

This was the most difficult of the extended response questions for the candidates. Finding s and t correctly in part (b) was difficult, with many confused between writing appropriate probabilities on a single branch compared to at the final end of a multiple branch. Many candidates had no idea what to write for a probability distribution and those who did often had probabilities that did not sum to 1. Candidates who wrote a probability distribution often could correctly compute the expected value. The final part was the most challenging, but some good answers were seen. The most common error was not recognizing that there were two different ways of winning.

This was the most difficult of the extended response questions for the candidates. Finding s and t correctly in part (b) was difficult, with many confused between writing appropriate probabilities on a single branch compared to at the final end of a multiple branch. Many candidates had no idea what to write for a probability distribution and those who did often had probabilities that did not sum to 1. Candidates who wrote a probability distribution often could correctly compute the expected value. The final part was the most challenging, but some good answers were seen. The most common error was not recognizing that there were two different ways of winning.

This was the most difficult of the extended response questions for the candidates. Finding s and t correctly in part (b) was difficult, with many confused between writing appropriate probabilities on a single branch compared to at the final end of a multiple branch. Many candidates had no idea what to write for a probability distribution and those who did often had probabilities that did not sum to 1. Candidates who wrote a probability distribution often could correctly compute the expected value. The final part was the most challenging, but some good answers were seen. The most common error was not recognizing that there were two different ways of winning.

Find the probability that

  (i)     none of the marbles are green;

  (ii)     exactly one marble is green.

Find the expected number of green marbles drawn from the jar.

(i)     Write down the probability that the marble is drawn from jar B.

(ii)     Given that the marble was drawn from jar B, write down the probability that it is red.

Given that the marble is red, find the probability that it was drawn from jar A.

(i)     attempt to find \({\rm{P(red)}} \times {\rm{P(red)}}\)     (M1)

eg   \(\frac{3}{8} \times \frac{2}{7}\) , \(\frac{3}{8} \times \frac{3}{8}\) , \(\frac{3}{8} \times \frac{2}{8}\)

\({\text{P(none green)}} = \frac{6}{{56}}\) \(\left( { = \frac{3}{{28}}} \right)\)     A1     N2

(ii)     attempt to find \({\rm{P(red)}} \times {\rm{P(green)}}\)     (M1)

eg   \(\frac{5}{8} \times \frac{3}{7}\) , \(\frac{3}{8} \times \frac{5}{8}\) , \(\frac{{15}}{{56}}\)

recognizing two ways to get one red, one green     (M1)

eg   \(2{\rm{P}}(R) \times {\rm{P}}(G)\) , \(\frac{5}{8} \times \frac{3}{7} + \frac{3}{8} \times \frac{5}{7}\) , \(\frac{3}{8} \times \frac{5}{8} \times 2\)

\({\text{P(exactly one green)}} = \frac{{30}}{{56}}\) \(\left( { = \frac{{15}}{{28}}} \right)\)     A1     N2

[5 marks]

\({\text{P(both green)}} = \frac{{20}}{{56}}\) (seen anywhere)     (A1)

correct substitution into formula for \({\rm{E}}(X)\)     A1

eg   \(0 \times \frac{6}{{56}} + 1 \times \frac{{30}}{{56}} + 2 \times \frac{{20}}{{56}}\) , \(\frac{{30}}{{64}} + \frac{{50}}{{64}}\)

expected number of green marbles is \(\frac{{70}}{{56}}\) \(\left( { = \frac{5}{4}} \right)\)     A1     N2

[3 marks]

(i)     \({\text{P(jar B)}} = \frac{4}{6}\) \(\left( { = \frac{2}{3}} \right)\)     A1     N1

(ii)     \({\text{P(red| jar B)}} = \frac{6}{8}\) \(\left( { = \frac{3}{4}} \right)\)     A1     N1

[2 marks]

recognizing conditional probability     (M1)

eg   \({\rm{P}}(A|R)\) , \(\frac{{{\text{P(jar A and red)}}}}{{{\rm{P(red)}}}}\) , tree diagram

attempt to multiply along either branch (may be seen on diagram)     (M1)

eg   \({\text{P(jar A and red)}} = \frac{1}{3} \times \frac{3}{8}\) \(\left( { = \frac{1}{8}} \right)\)

attempt to multiply along other branch     (M1)

eg   \({\text{P(jar B and red)}} = \frac{2}{3} \times \frac{6}{8}\) \(\left( { = \frac{1}{2}} \right)\)

adding the probabilities of two mutually exclusive paths     (A1)

eg   \({\rm{P(red)}} = \frac{1}{3} \times \frac{3}{8} + \frac{2}{3} \times \frac{6}{8}\)

correct substitution

eg   \({\text{P(jar A|red)}} = \frac{{\frac{1}{3} \times \frac{3}{8}}}{{\frac{1}{3} \times \frac{3}{8} + \frac{2}{3} \times \frac{6}{8}}}\) , \(\frac{{\frac{1}{8}}}{{\frac{5}{8}}}\)     A1

\({\text{P(jar A|red)}} = \frac{1}{5}\)     A1     N3

[6 marks]

Many candidates correctly found the probability of selecting no green marbles in two draws, although some candidates treated the second draw as if replacing the first. When finding the probability for exactly one green marble, candidates often failed to recognize two pathways for selecting one of each color.

Few candidates understood the concept of expected value in this context, often leaving this blank or treating as if a binomial experiment. Successful candidates often made a distribution table before making the final calculation.

Most candidates answered part (c) correctly. However, many overcomplicated (c)(ii) by using the conditional probability formula. Those with a clear understanding of the concept easily followed the “write down” instruction.

Only a handful of candidates correctly applied conditional probability to find \(P(A|R)\) in part (d). While some wrote down the formula, or drew a tree diagram, few correctly calculated \(P(red) = \frac{5}{8}\) . A common error was to combine the marbles in the two jars to give \(P(red) = \frac{9}{16}\) .

Find the value of \(p\).

Find the probability that Adam will travel by car and be late for school.

Find the probability that Adam will be late for school.

Given that Adam is late for school, find the probability that he travelled by car.

Adam will go to school three times next week.

Find the probability that Adam will be late exactly once.

correct working     (A1)

eg\(\;\;\;1 - \frac{1}{6}\)

\(p = \frac{5}{6}\)     A1     N2

[2 marks]

multiplying along correct branches     (A1)

eg\(\;\;\;\frac{1}{2} \times \frac{1}{6}\)

\({\text{P}}(C \cap L) = \frac{1}{{12}}\)     A1     N2

[2 marks]

multiplying along the other branch     (M1)

eg\(\;\;\;\frac{1}{2} \times \frac{1}{3}\)

adding probabilities of their \(2\) mutually exclusive paths     (M1)

eg\(\;\;\;\frac{1}{2} \times \frac{1}{6} + \frac{1}{2} \times \frac{1}{3}\)

correct working     (A1)

eg\(\;\;\;\frac{1}{{12}} + \frac{1}{6}\)

\({\text{P}}(L) = \frac{3}{{12}}\;\;\;\left( { = \frac{1}{4}} \right)\)     A1     N3

[4 marks]

recognizing conditional probability (seen anywhere)     (M1)

eg\(\;\;\;{\text{P}}(C|L)\)

correct substitution of their values into formula     (A1)

eg\(\frac{{\frac{1}{{12}}}}{{\frac{3}{{12}}}}\)

\({\text{P}}(C|L) = \frac{1}{3}\)     A1     N2

[3 marks]

valid approach     (M1)

eg  \(X \sim B\) \(\left({3,{\text{ }}\frac{1}{4}} \right),{\text{ }}\left({\frac{1}{4}} \right){\left({\frac{3}{4}} \right)^2},{\text{ }}\left( {\begin{array}{*{20}{c}}3 \\1\end{array}} \right)\), three ways it could happen

correct substitution     (A1)

eg  \(\;\;\;\left({\begin{array}{*{20}{c}}3\\1\end{array}}\right){\left({\frac{1}{4}}\right)^1}{\left({\frac{3}{4}} \right)^2},{\text{ }}\frac{1}{4}\times\frac{3}{4}\times\frac{3}{4} + \frac{3}{4} \times\frac{1}{4}\times\frac{3}{4} + \frac{3}{4}\times\frac{3}{4}\times\frac{1}{4}\)

correct working     (A1)

eg\(\;\;\;3\left( {\frac{1}{4}} \right)\left( {\frac{9}{{16}}} \right),{\text{ }}\frac{9}{{64}} + \frac{9}{{64}} + \frac{9}{{64}}\)

\(\frac{{27}}{{64}}\)     A1     N2

[4 marks]

Total [15 marks]

Parts (a) and (b) of this question were answered correctly by nearly all candidates, and the majority earned full marks on part (c), as well. Unfortunately, there were a number of candidates who made arithmetic errors when multiplying or adding fractions.

Parts (a) and (b) of this question were answered correctly by nearly all candidates, and the majority earned full marks on part (c), as well. Unfortunately, there were a number of candidates who made arithmetic errors when multiplying or adding fractions. Candidates were not as successful in parts (d) and (e) of this question. Although many knew that conditional probability was necessary in part (d), many did not know to use their values from parts (b) and (c), and started from scratch with brand new, and often incorrect, calculations for the numerator and denominator. A majority of candidates did not recognize that binomial probability was needed in part (e), not realizing that there were three ways for Adam to be "late exactly once". A very common incorrect solution to part (e) was \(\frac{1}{4} \times \frac{3}{4} \times \frac{3}{4} = \frac{9}{{64}}\).

Parts (a) and (b) of this question were answered correctly by nearly all candidates, and the majority earned full marks on part (c), as well. Unfortunately, there were a number of candidates who made arithmetic errors when multiplying or adding fractions. Candidates were not as successful in parts (d) and (e) of this question. Although many knew that conditional probability was necessary in part (d), many did not know to use their values from parts (b) and (c), and started from scratch with brand new, and often incorrect, calculations for the numerator and denominator. A majority of candidates did not recognize that binomial probability was needed in part (e), not realizing that there were three ways for Adam to be "late exactly once". A very common incorrect solution to part (e) was \(\frac{1}{4} \times \frac{3}{4} \times \frac{3}{4} = \frac{9}{{64}}\).

Parts (a) and (b) of this question were answered correctly by nearly all candidates, and the majority earned full marks on part (c), as well. Unfortunately, there were a number of candidates who made arithmetic errors when multiplying or adding fractions. Candidates were not as successful in parts (d) and (e) of this question. Although many knew that conditional probability was necessary in part (d), many did not know to use their values from parts (b) and (c), and started from scratch with brand new, and often incorrect, calculations for the numerator and denominator. A majority of candidates did not recognize that binomial probability was needed in part (e), not realizing that there were three ways for Adam to be "late exactly once". A very common incorrect solution to part (e) was \(\frac{1}{4} \times \frac{3}{4} \times \frac{3}{4} = \frac{9}{{64}}\).

Parts (a) and (b) of this question were answered correctly by nearly all candidates, and the majority earned full marks on part (c), as well. Unfortunately, there were a number of candidates who made arithmetic errors when multiplying or adding fractions. Candidates were not as successful in parts (d) and (e) of this question. Although many knew that conditional probability was necessary in part (d), many did not know to use their values from parts (b) and (c), and started from scratch with brand new, and often incorrect, calculations for the numerator and denominator. A majority of candidates did not recognize that binomial probability was needed in part (e), not realizing that there were three ways for Adam to be "late exactly once". A very common incorrect solution to part (e) was \(\frac{1}{4}\times\frac{3}{4} \times\frac{3}{4} = \frac{9}{{64}}\).

Bag A contains three white balls and four red balls. Two balls are chosen at random without replacement.

(i)     Copy and complete the following tree diagram.

(ii)    Find the probability that two white balls are chosen.

Bag A contains three white balls and four red balls. Two balls are chosen at random without replacement.

Bag B contains four white balls and three red balls. When two balls are chosen at random without replacement from bag B, the probability that they are both white is \(\frac{2}{7}\) .

A standard die is rolled. If 1 or 2 is obtained, two balls are chosen without replacement from bag A, otherwise they are chosen from bag B.

Find the probability that the two balls are white.

Bag A contains three white balls and four red balls. Two balls are chosen at random without replacement.

Bag B contains four white balls and three red balls. When two balls are chosen at random without replacement from bag B, the probability that they are both white is \(\frac{2}{7}\) . 

A standard die is rolled. If 1 or 2 is obtained, two balls are chosen without replacement from bag A, otherwise they are chosen from bag B.

Given that both balls are white, find the probability that they were chosen from bag A.

(i)

\(\frac{4}{6},\frac{3}{6}{\rm{and}}\frac{3}{6}\left( {\frac{2}{3},\frac{1}{2}{\rm{and}}\frac{1}{2}} \right)\)    A1A1A1     N3

(ii) multiplying along the correct branches (may be seen on diagram)     (A1)

e.g. \(\frac{3}{7} \times \frac{2}{6}\)

\(\frac{6}{{42}}\left( { = \frac{1}{7}} \right)\)     A1     N2

[5 marks]

\({\rm{P(bag A) = }}\frac{2}{6}\left( { = \frac{1}{3}} \right)\) , \({\rm{P(bag B) = }}\frac{4}{6}\left( { = \frac{2}{3}} \right)\) (seen anywhere)     (A1)(A1)

appropriate approach    (M1)

e.g. \({\rm{P(}}WW \cap A) + {\rm{P}}(WW \cap B)\)

correct calculation     A1

e.g. \(\frac{1}{3} \times \frac{1}{7} + \frac{2}{3} \times \frac{2}{7}\) , \(\frac{2}{{42}} + \frac{8}{{42}}\)

\({\rm{P}}(2W) = \frac{{60}}{{252}}\left( { = \frac{5}{{21}}} \right)\)    A1     N3

[5 marks]

recognizing conditional probability     (M1)

e.g. \(\frac{{{\rm{P}}(A \cap B)}}{{{\rm{P}}(B)}}\) , \({\rm{P}}(A|WW) = \frac{{{\rm{P}}(WW \cap A)}}{{{\rm{P}}(WW)}}\)

correct numerator     (A1)

e.g. \({\rm{P}}(A \cap WW) = \frac{6}{{42}} \times \frac{2}{6},\frac{1}{{21}}\)

correct denominator     (A1)

e.g. \(\frac{6}{{252}},\frac{5}{{21}}\)

probability \(\frac{{84}}{{420}}\left( { = \frac{1}{5}} \right)\)    A1     N3

[4 marks]

Part (a) of this question was answered correctly by the large majority of candidates. There were some who did not follow the instruction to copy and complete the tree diagram on their separate paper, and simply filled in the blanks on the exam paper.

In part (b), many candidates struggled with finding the compound probability, and did not use the provided information in the appropriate manner. Quite a few candidates seemed to be confused about when they should add the probabilities or when they should multiply.

In part (c), many recognized that the question dealt with conditional probability, and many tried to use the formula from the information booklet, but failed to realize that they had already found the required values for the numerator and denominator in their working for part (b).

Throughout this question, it was discouraging to see the large number of candidates making arithmetic errors. There were a surprising number of candidates who multiplied fractions incorrectly, or found an incorrect value for simple multiplication such as \(2 \times 4 = 6\) or \(6 \times 7 = 43\) .

Find the number of students who completed the task in less than 45 minutes.

Find the number of students who took between 35 and 45 minutes to complete the task.

Given that 50 students take less than k minutes to complete the task, find the value of \(k\).

attempt to find number who took less than 45 minutes     (M1)

eg     line on graph (vertical at approx 45, or horizontal at approx 70)

70 students   (accept 69)     A1     N2

[2 marks]

55 students completed task in less than 35 minutes     (A1)

subtracting their values     (M1)

eg     70 – 55

15 students     A1     N2

[3 marks]

correct approach     (A1)

eg     line from y-axis on 50

\(k = 33\)     A1     N2

[2 marks]

Write down an expression for \({\text{P}}(A \cap B)\) .

Given that \({\text{P}}(A \cup B) = 0.8\) ,

(i)     find x ;

(ii)    find \({\text{P}}(A \cap B)\) .

Hence, explain why A and B are not mutually exclusive.

\({\text{P}}(A \cap B) = {\text{P}}(A) \times {\text{P}}(B)( = 0.6x)\)    A1     N1

[1 mark]

(i) evidence of using \({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B) - {\text{P}}(A){\text{P}}(B)\)     (M1)

correct substitution     A1

e.g. \(0.8 = 0.6 + x - 0.6x\) , \(0.2 = 0.4x\)

\(x = 0.5\)    A1     N2

(ii) \({\text{P}}(A \cap B) = 0.3\)     A1     N1

[4 marks]

valid reason, with reference to \({\text{P(}}A \cap B)\)     R1     N1

e.g. \({\text{P(}}A \cap B) \ne 0\)

[1 mark]

This question was well done by most candidates.

This question was well done by most candidates. When errors were made, candidates confused the terms "independent" and "mutually exclusive" and did not subtract the intersection when finding \({\text{P}}(A \cup B)\) .

Candidates should also be aware of the command term "hence" used in part (c) where they were expected to provide a reason that involved \({\text{P}}(A \cap B)\) from their work in part (b). It seemed that many turned to the formula in the booklet instead of considering the conceptual meaning of the term.

Find \(p\).

Find \({\text{E}}(X)\).

summing probabilities to 1     (M1)

eg,\(\;\;\;\sum { = 1,{\text{ }}3 + 4 + 2 + x = 10} \)

correct working     (A1)

\(\frac{3}{{10}} + \frac{4}{{10}} + \frac{2}{{10}} + p = 1,{\text{ }}p = 1 - \frac{9}{{10}}\)

\(p = \frac{1}{{10}}\)     A1     N3

[3 marks]

correct substitution into formula for \({\text{E}}(X)\)     (A1)

eg\(\;\;\;0\left( {\frac{3}{{10}}} \right) +  \ldots  + 3(p)\)

correct working     (A1)

eg\(\;\;\;\frac{4}{{10}} + \frac{4}{{10}} + \frac{3}{{10}}\)

\({\text{E}}(X) = \frac{{11}}{{10}}\;\;\;(1.1)\)     A1     N2

[3 marks]

Total [6 marks]

Most candidates were able to find \(p\), however expectation emerged as surprisingly more difficult. Quite often \({\text{E}}(X)/4\) was found or candidates wrote the formula with no further work.

Most candidates were able to find \(p\), however expectation emerged as surprisingly more difficult. Quite often \({\text{E}}(X)/4\) was found or candidates wrote the formula with no further work.

One student is selected at random.

(i)     Calculate the probability that the student is a male or is a tennis player.

(ii)    Given that the student selected is female, calculate the probability that the student does not play football.

Two students are selected at random. Calculate the probability that neither student plays football.

(i) correct calculation     (A1)

e.g. \(\frac{9}{{20}} + \frac{5}{{20}} - \frac{2}{{20}}\) , \(\frac{{4 + 2 + 3 + 3}}{{20}}\)

\({\text{P(male or tennis)}} = \frac{{12}}{{20}}\)     A1     N2

(ii) correct calculation     (A1)

e.g. \(\frac{6}{{20}} \div \frac{{11}}{{20}}\) , \(\frac{{3 + 3}}{{11}}\)

\({\text{P(not football|female)}} = \frac{6}{{11}}\)     A1     N2

[4 marks]

\({\text{P(first not football)}} = \frac{{11}}{{20}}\) , \({\text{P(second not football)}} = \frac{{10}}{{19}}\)     A1

\({\text{P(neither football)}} = \frac{{11}}{{20}} \times \frac{{10}}{{19}}\)     A1

\({\text{P(neither football)}} = \frac{{110}}{{380}}\)     A1     N1

[3 marks]

Many candidates had difficulty with this question, usually as a result of seeking to solve the problem by formula instead of looking carefully at the table frequencies.

A very common error in part (b) was to assume identical probabilities for each selection instead of dependent probabilities where there is no replacement.

Find AB .

Given that \({\boldsymbol{X}} - 2{\boldsymbol{A}} = {\boldsymbol{B}}\), find X.

evidence of multiplying     (M1)

e.g. one correct element, \((0 \times - 4) + (3 \times 5)\)

\({\boldsymbol{AB}} = \left( {\begin{array}{*{20}{c}}
{15}&3\\
{28}&4
\end{array}} \right)\)     A2     N3

Note: Award A1 for three correct elements.

[3 marks]

finding \(2{\boldsymbol{A}} = \left( {\begin{array}{*{20}{c}}
0&6\\
{ - 4}&8
\end{array}} \right)\)     (A1)

adding 2\({\boldsymbol{A}}\) to both sides (may be seen first)     (M1)

e.g. \({\boldsymbol{X}} = {\boldsymbol{B}} +2{\boldsymbol{A}}\)

 \({\boldsymbol{X}} = \left( {\begin{array}{*{20}{c}}
{ - 4}&6\\
1&9
\end{array}} \right)\)     A1     N2

[3 marks]

The large majority of candidates answered this question successfully. There were only a small number of candidates who seemed to have never worked with matrices before. Occasionally a candidate would incorrectly approach part (b) by finding an inverse of matrix A.

The large majority of candidates answered this question successfully. There were only a small number of candidates who seemed to have never worked with matrices before. Occasionally a candidate would incorrectly approach part (b) by finding an inverse of matrix A.

Write down \({\rm{P}}(X = 2)\) .

Show that \(k = 3\) .

Find \({\rm{E}}(X)\) .

\({\rm{P}}(X = 2) = \frac{4}{{14}}\) \(\left( { = \frac{2}{7}} \right)\)     A1     N1

[1 mark]

\({\rm{P}}(X = 1) = \frac{1}{{14}}\)     (A1)

\({\rm{P}}(X = k) = \frac{{{k^2}}}{{14}}\)     (A1)

setting the sum of probabilities \( = 1\)     M1

e.g. \(\frac{1}{{14}} + \frac{4}{{14}} + \frac{{{k^2}}}{{14}} = 1\) , \(5 + {k^2} = 14\)

\({k^2} = 9\) (accept \(\frac{{{k^2}}}{{14}} = \frac{9}{{14}}\) )     A1

\(k = 3\)     AG     N0

[4 marks]

correct substitution into \({\rm{E}}(X) = \sum {x{\rm{P}}(X = x)} \)     A1

e.g. \(1\left( {\frac{1}{{14}}} \right) + 2\left( {\frac{4}{{14}}} \right) + 3\left( {\frac{9}{{14}}} \right)\)

\({\rm{E}}(X) = \frac{{36}}{{14}}\) \(\left( { = \frac{{18}}{7}} \right)\)    A1     N1

[2 marks]

Although many candidates were successful in working with the probability function, students had difficulty following the "show that" instruction of this question. Many substituted \(k = 3\) and worked backwards to show that the sum of probabilities is 1. Some would argue that \(k = 4\) does not work, but were unable to give a complete justification for \(k = 3\) . A good number of students seemed unprepared to find an expected value. Many candidates wrote a formula and did not know what to do with it, while others divided \({\rm{E}}(X)\) by 3 or by 6, which confuses the concept of a mean in a probability distribution with the more common understanding.

Although many candidates were successful in working with the probability function, students had difficulty following the "show that" instruction of this question. Many substituted \(k = 3\) and worked backwards to show that the sum of probabilities is 1. Some would argue that \(k = 4\) does not work, but were unable to give a complete justification for \(k = 3\) .

A good number of students seemed unprepared to find an expected value. Many candidates wrote a formula and did not know what to do with it, while others divided \({\rm{E}}(X)\) by 3 or by 6, which confuses the concept of a mean in a probability distribution with the more common understanding.

Find the value of k .

Find \({\text{E}}(X)\) .

evidence of summing to 1     (M1)

e.g. \(\sum\limits_{}^{} {p = 1{\text{, }}0.3 + k + 2k + 0.1 = 1} \)

correct working     (A1)

e.g. \(0.4 + 3k{\text{, }}3k = 0.6\)

\(k = 0.2\)    A1     N2

[3 marks]

correct substitution into formula \({\text{E}}(X)\)      (A1)

e.g. \(0(0.3) + 2(k) + 5(2k) + 9(0.1){\text{, }}12k + 0.9\)

correct working 

e.g. \(0(0.3) + 2(0.2) + 5(0.4) + 9(0.1){\text{, }}0.4 + 2.0 + 0.9\)     (A1)

\({\text{E}}(X)\)= 3.3     A1     N2

[3 marks]

Overall, this question was very well done. A few candidates left this question blank, or used methods which would indicate they were unfamiliar with discrete random variables. In part (b), there were a good number of candidates who set up their work correctly, but then had trouble adding or multiplying decimals without a calculator. A common type of error for these candidates was \(5(0.4) = 0.2\) .

Overall, this question was very well done. A few candidates left this question blank, or used methods which would indicate they were unfamiliar with discrete random variables. In part (b), there were a good number of candidates who set up their work correctly, but then had trouble adding or multiplying decimals without a calculator. A common type of error for these candidates was \(5(0.4) = 0.2\) .

Find the value of \(p\);

Find the value of \(q\).

A girl is selected at random. Find the probability that she takes economics but not history.

valid approach     (M1)

eg\(\,\,\,\,\,\)\(p + 3 = 13,{\text{ }}13 - 3\)

\(p = 10\)     A1     N2

[2 marks]

valid approach     (M1)

eg\(\,\,\,\,\,\)\(p + 3 + 5 + q = 20,{\text{ }}10 - 10 - 8\)

\(q = 2\)     A1     N2

[2 marks]

valid approach     (M1)

eg\(\,\,\,\,\,\)\(20 - p - q - 3,{\text{ }}1 - \frac{{15}}{{20}},{\text{ }}n(E \cap H') = 5\)

\(\frac{5}{{20}}\,\,\,\left( {\frac{1}{4}} \right)\)     A1     N2

[2 marks]

Find the probability that Ann wins on her first roll.

(i)     The probability that Ann wins on her third roll is \(\frac{5}{8} \times \frac{4}{8} \times p \times q\ \times \frac{3}{8}\).

Write down the value of \(p\) and of \(q\).

(ii)     The probability that Ann wins on her tenth roll is \(\frac{3}{8}{r^k}\) where \(r \in \mathbb{Q},{\text{ }}k \in \mathbb{Z}\).

Find the value of \(r\) and of \(k\).

Find the probability that Ann wins the game.

recognizing Ann rolls green     (M1)

eg\(\;\;\;{\text{P(G)}}\)

\(\frac{3}{8}\)     A1     N2

[2 marks]

(i)     \(p = \frac{4}{8},{\text{ }}q = \frac{5}{8}\) or \(q = \frac{4}{8},{\text{ }}p = \frac{5}{8}\)     A1A1     N2

(ii)     recognizes Ann and Bob lose \(9\) times     (M1)

eg\(\;\;\;\(\overbrace {{A_L}{B_\(\overbrace {{A_L}{B_ \ldots \(\overbrace {{A_L}{B_{\text{ 9 times, }}\underbrace {\left( {\frac{5}{8} \times \frac{4}{8}} \right) \times  \ldots  \times \left( {\frac{5}{8} \times \frac{4}{8}} \right)}_{{\text{9 times}}}\)

\(k = 9\;\;\;\)(seen anywhere)     A1     N2

correct working     (A1)

eg\(\;\;\;{\left( {\frac{5}{8} \times \frac{4}{8}} \right)^9} \times \frac{3}{8},{\text{ }}\left( {\frac{5}{8} \times \frac{4}{8}} \right) \times  \ldots  \times \left( {\frac{5}{8} \times \frac{4}{8}} \right) \times \frac{3}{8}\)

\(r = \frac{{20}}{{64}}\;\;\;\left( { = \frac{5}{{16}}} \right)\)     A1     N2

[6 marks]

recognize the probability is an infinite sum     (M1)

eg\(\;\;\;\)Ann wins on her \({{\text{1}}^{{\text{st}}}}\) roll or \({{\text{2}}^{{\text{nd}}}}\) roll or \({{\text{3}}^{{\text{rd}}}}\) roll…, \({S_\infty }\)

recognizing GP     (M1)

\({u_1} = \frac{3}{8}\;\;\;\)(seen anywhere)     A1

\(r = \frac{{20}}{{64}}\;\;\;\)(seen anywhere)     A1

correct substitution into infinite sum of GP     A1

eg\(\;\;\;\frac{{\frac{3}{8}}}{{1 - \frac{5}{{16}}}},{\text{ }}\frac{3}{8}\left( {\frac{1}{{1 - \left( {\frac{5}{8} \times \frac{4}{8}} \right)}}} \right),{\text{ }}\frac{1}{{1 - \frac{5}{{16}}}}\)

correct working     (A1)

eg\(\;\;\;\frac{{\frac{3}{8}}}{{\frac{{11}}{{16}}}},{\text{ }}\frac{3}{8} \times \frac{{16}}{{11}}\)

\({\text{P (Ann wins)}} = \frac{{48}}{{88}}\;\;\;\left( { = \frac{6}{{11}}} \right)\)     A1     N1

[7 marks]

Total [15 marks]

Some teachers’ comments suggested that the word ‘loses’ in the diagram was misleading. But candidate scripts did not indicate any adverse effect.

a)     Very well answered.

b)     i) Probabilities \(p\) and \(q\) were typically found correctly. ii) Fewer candidates identified the common ratio and number of rolls correctly.

Few candidates recognized that this was an infinite geometric sum although some did see that a geometric progression was involved.

Some teachers’ comments suggested that the word ‘loses’ in the diagram was misleading, But candidate scripts did not indicate any adverse effect.

a)     Very well answered.

b)     i) Probabilities \(p\) and \(q\) were typically found correctly. ii) Fewer candidates identified the common ratio and number of rolls correctly.

Few candidates recognized that this was an infinite geometric sum although some did see that a geometric progression was involved.

Some teachers’ comments suggested that the word ‘loses’ in the diagram was misleading, But candidate scripts did not indicate any adverse effect.

a)     Very well answered.

b)     i) Probabilities \(p\) and \(q\) were typically found correctly. ii) Fewer candidates identified the common ratio and number of rolls correctly.

Few candidates recognized that this was an infinite geometric sum although some did see that a geometric progression was involved.

Each value in the data set has 10 added to it. Write down the value of

(i)     the new mean;

(ii)    the new standard deviation.

Each value in the original data set is multiplied by 10.

(i)     Write down the value of the new mean.

(ii)    Find the value of the new variance.

(i) new mean is \(20 + 10 = 30\)     A1     N1

(ii) new sd is 6     A1     N1

[2 marks]

(i) new mean is \(20 \times 10 = 200\)    A1     N1

(ii) METHOD 1

variance is 36     A1

new variance is \(36 \times 100 = 3600\)     A1     N2

METHOD 2

new sd is 60     A1

new variance is \({60^2} = 3600\)     A1     N2

[3 marks]

Write down the median score.

The range of the scores is 47 marks, and the interquartile range is 22 marks.

Find the value of

(i)     \(c\);

(ii)     \(d\).

60     A1     N1

(i)     valid approach     (M1)

eg \(\max - \min = {\rm{range}},c = 40 + 47\)

\(c = 87\)     A1     N2

(ii)     valid approach     (M1)

eg\(\;\;\;Q3 - Q1 = IQR,{\text{ }}74 - 22\)

\(d = 52\)     A1     N2

Find n.

Write down the value of the new mean.

Find the value of the new variance.

correct approach      (A1)

eg  \(\frac{{800}}{n} = 20\)

40      A1 N2

[2 marks]

200   A1 N1

[1 mark]

METHOD 1

recognizing variance = σ ²      (M1)

eg  3² = 9

correct working to find new variance      (A1)

eg  σ ² × 10², 9 × 100

900      A1 N3

METHOD 2

new standard deviation is 30      (A1)

recognizing variance = σ ²      (M1)

eg 3² = 9, 30²

900      A1 N3

[3 marks]

Find the mean.

(i)     Write down the value of the new mean.

(ii)     Find the value of the new variance.

correct approach     (A1)

eg\(\,\,\,\,\,\)\(\frac{{60}}{{10}}\)

\({\text{mean}} = 6\)    A1     N2

(i)     new mean \( = 24\)     A1     N1

(ii)     valid approach     (M1)

eg\(\,\,\,\,\,\)\({\text{variance}} \times {(4)^2},{\text{ }}3 \times 16\), new standard deviation \( = 4\sqrt 3 \)

new variance \( = 48\)     A1     N2

[3 marks]

While most candidates were able to answer part (a) of this question correctly, they were not as successful in part (b). It seems that the item on the Maths SL syllabus dealing with the "effect of constant changes to the original data" was skipped over in many schools.

While most candidates were able to answer part (a) of this question correctly, they were not as successful in part (b). It seems that the item on the Maths SL syllabus dealing with the "effect of constant changes to the original data" was skipped over in many schools.

Find

(i)     \({\rm{P}}(E \cap F)\) ;

(ii)    \({\rm{P}}(F)\) .

Find the probability that

(i)     José misses his bus and is not late for school;

(ii)    José missed his bus, given that he is late for school.

The cost for each day that José catches the bus is 3 euros. José goes to school on Monday and Tuesday.

Copy and complete the probability distribution table.

The cost for each day that José catches the bus is 3 euros. José goes to school on Monday and Tuesday.

Find the expected cost for José for both days.

(i) \(\frac{7}{{24}}\)     A1     N1

(ii) evidence of multiplying along the branches     (M1)

e.g. \(\frac{2}{3} \times \frac{5}{8}\) , \(\frac{1}{3} \times \frac{7}{8}\)

adding probabilities of two mutually exclusive paths     (M1)

e.g. \(\left( {\frac{1}{3} \times \frac{7}{8}} \right) + \left( {\frac{2}{3} \times \frac{3}{8}} \right)\) , \(\left( {\frac{1}{3} \times \frac{1}{8}} \right) + \left( {\frac{2}{3} \times \frac{5}{8}} \right)\)

\({\rm{P}}(F) = \frac{{13}}{{24}}\)     A1     N2

[4 marks]

(i) \(\frac{1}{3} \times \frac{1}{8}\)     (A1)

\(\frac{1}{{24}}\)     A1

(ii) recognizing this is \({\rm{P}}(E|F)\)     (M1)

e.g. \(\frac{7}{{24}} \div \frac{{13}}{{24}}\)

\(\frac{{168}}{{312}}\) \(\left( { = \frac{7}{{13}}} \right)\)     A2     N3

[5 marks]

     A2A1     N3

[3 marks]

correct substitution into \({\rm{E}}(X)\) formula     (M1)

e.g. \(0 \times \frac{1}{9} + 3 \times \frac{4}{9} + 6 \times \frac{4}{9}\) , \(\frac{{12}}{9} + \frac{{24}}{9}\)

\({\rm{E}}(X) = 4\) (euros)     A1     N2

[2 marks]

Candidates generally handled some or all of parts (a) and (b) well. Errors included adding probabilities along branches and trying to use the union formula from the information booklet.

Candidates generally handled some or all of parts (a) and (b) well. Errors included adding probabilities along branches and trying to use the union formula from the information booklet. On part (b)(ii), many candidates knew that they were supposed to use some type of conditional probability but did not know how to find \({\rm{P}}(E|F)\) . Many candidates made errors working with fractions. Some candidates who missed part (a)(ii) were able to earn follow-through credit on part (b)(ii).

Many candidates had difficulty completing the probability distribution table. While the common error of finding the probability for \(x = 3\) as \(\frac{2}{9}\) was understandable as the candidate did not appreciate that there were two ways of paying three euros, it was disappointing that these candidates often correctly found \({\rm{P}}(X = 4)\) as \(\frac{4}{9}\) and did not note that the probabilities failed to sum to one. These candidates could not earn full follow-through marks on their expected value calculation in part (d). Some candidates did use the probabilities summing to one with incorrect probabilities in part (c); these candidates often earned full follow-through marks in part (d), as a majority of candidates knew the method for finding expected value.

Many candidates had difficulty completing the probability distribution table. While the common error of finding the probability for \(x = 3\) as \(\frac{2}{9}\) was understandable as the candidate did not appreciate that there were two ways of paying three Euros, it was disappointing that these candidates often correctly found \({\rm{P}}(X) = 4\) as \(\frac{4}{9}\) and did not note that the probabilities failed to sum to one. These candidates could not earn full follow-through marks on their expected value calculation in part (d). Some candidates did use the probabilities summing to one with incorrect probabilities in part (c); these candidates often earned full follow-through marks in part (d), as a majority of candidates knew the method for finding expected value.

(i)     Write down the value of s .

(ii)    Find the value of q .

(iii)   Write down the value of p and of r .

(i)     A student is selected at random. Given that the student takes music, write down the probability the student takes art.

(ii)    Hence, show that taking music and taking art are not independent events.

Two students are selected at random, one after the other. Find the probability that the first student takes only music and the second student takes only art.

(i) \(s = 1\)     A1     N1

(ii) evidence of appropriate approach     (M1)

e.g. \(21 - 16\) , \(12 + 8 - q = 15\)

\(q = 5\)     A1     N2

(iii) \(p = 7\) , \(r = 3\)     A1A1     N2

[5 marks]

(i) \({\rm{P(art|music)}} = \frac{5}{8}\)     A2     N2

(ii) METHOD 1

\({\rm{P(art)}} = \frac{{12}}{{16}}\) \(\left( { = \frac{3}{4}} \right)\)     A1

evidence of correct reasoning     R1

e.g. \(\frac{3}{4} \ne \frac{5}{8}\)

the events are not independent     AG     N0

METHOD 2

\({\rm{P(art)}} \times {\rm{P(music)}} = \frac{{96}}{{256}}\) \(\left( { = \frac{3}{8}} \right)\)     A1

evidence of correct reasoning     R1

e.g. \(\frac{{12}}{{16}} \times \frac{8}{{16}} \ne \frac{5}{{16}}\)

the events are not independent     AG     N0

[4 marks]

\({\text{P(first takes only music)}} = \frac{3}{{16}}\) (seen anywhere)     A1

\({\text{P(second takes only art)}} = \frac{7}{{15}}\) (seen anywhere)     A1

evidence of valid approach     (M1)

e.g. \(\frac{3}{{16}} \times \frac{7}{{15}}\)

\({\text{P(music and art)}} = \frac{{21}}{{240}}\) \(\left( { = \frac{7}{{80}}} \right)\)     A1     N2

[4 marks]

A majority of candidates found the values in the Venn diagram easily. Common errors include giving \(s = 16\) , and also neglecting s in finding \(q = 4\) (e.g. \(12 + 8 - 16\) ) . Some interpreted the values as probabilities, despite the question explicitly stating that p, q, r and s represent numbers of students. Occasionally the values for p and r were misinterpreted as being inclusive of q. Follow-through marks were often earned in subsequent parts for such cases.

For (b), rather than think of the situations conceptually, most candidates reached for the formula for conditional probability, with mixed results. Few candidates considered that independence means \({\rm{P}}(A|M) = {\rm{P}}(A)\) . Most applied \({\rm{P}}(A \cap B) = {\rm{P}}(A) \times {\rm{P}}(B)\) , with many giving incomplete or incorrect calculations. Some candidates compared the wrong things and showed, for example, that \(\frac{5}{8} \ne \frac{3}{8}\) , which incorrectly compares \({\rm{P}}(A|M)\) with \({\rm{P}}(A \cap M)\) . Others stated that because there is an intersection, the events are independent, which is an insufficient explanation.

Part (c) was commonly answered as if there is replacement, with many candidates calculating \(\frac{3}{{16}} \times \frac{7}{{16}}\) . However, implicit in the phrasing "one after the other" is that there is no replacement.

Find the value of \(p\) and of \(q\) .

A girl is chosen at random.

  (i)     Find the probability that the time she takes is less than \(14\) minutes.

  (ii)     Find the probability that the time she takes is at least \(26\) minutes.

A girl is selected for the competition if she takes less than \(x\) minutes to complete the race.

Given that \(40\%\) of the girls are not selected,

  (i)     find the number of girls who are not selected;

  (ii)     find \(x\) .

Girls who are not selected, but took less than \(25\) minutes to complete the race, are allowed another chance to be selected. The new times taken by these girls are shown in the cumulative frequency diagram below.

(i)     Write down the number of girls who were allowed another chance.

(ii)     Find the percentage of the whole group who were selected.

attempt to find \(p\)     (M1)

eg   \(120 - 70\) , \(50 + 20 + x = 120\)

\(p = 50\)     A1     N2

attempt to find \(q\)     (M1)

eg   \(180 - 20\) , \(200 - 20 - 20\)

\(q = 160\)     A1     N2

[4 marks]

(i)     \(\frac{{70}}{{200}}\) \(\left( { = \frac{7}{{20}}} \right)\)     A1     N1

(ii)     valid approach     (M1)

eg   \(20 + 20\) , \(200 - 160\)

\(\frac{{40}}{{200}}\) \(\left( { = \frac{1}{5}} \right)\)     A1     N2

[3 marks]

(i)     attempt to find number of girls     (M1)

eg   \(0.4\), \(\frac{{40}}{{100}} \times 200\)

\(80\) are not selected    A1     N2

(ii)     \(120\) are selected     (A1)

\(x = 20\)     A1     N2

[4 marks]

(i)     \(30\) given second chance     A1     N1

(ii)     \(20\) took less than \(20\) minutes     (A1)

attempt to find their selected total (may be seen in \(\%\) calculation)     (M1)

eg   \(120 + 20\) \(( = 140)\) , \(120 + \) their answer from (i)

\(70\) (\(\%\))     A1     N3

[4 marks]

Overall, candidates were very successful in parts (a), (b) and (c) of this question. Most of the errors in these parts had to do with candidates not understanding terms such as "at least" or "less than".

Overall, candidates were very successful in parts (a), (b) and (c) of this question. Most of the errors in these parts had to do with candidates not understanding terms such as "at least" or "less than".

Overall, candidates were very successful in parts (a), (b) and (c) of this question. Most of the errors in these parts had to do with candidates not understanding terms such as "at least" or "less than".

Part (d) was quite challenging for candidates, who may not have read the question carefully and studied the values in the diagram. Many seemed confused by the idea that not all the girls who were given a second chance were selected. In part (d)(ii), many did not find the percentage of the whole group, but rather the percentage of the girls who were given a second chance.

(i)     Write down the value of \(q\).

(ii)     Find the value of \(p\).

Find \({\text{P}}(B)\).

(i)     \(q = 0.1\)     A1     N1

(ii)     appropriate approach     (M1)

eg\(\,\,\,\,\,\)\({\text{P}}(A) - q,{\text{ }}0.4 - 0.1\)

\(p = 0.3\)    A1     N2

[3 marks]

valid approach     (M1)

eg \(\,\,\,\,\,\)\({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B) - {\text{P}}(A \cap B),{\text{ P}}(A \cap B) + {\text{P}}(B \cap A')\)

correct values     (A1)

eg\(\,\,\,\,\,\)\(0.8 = 0.4 + {\text{P}}(B) - 0.1,{\text{ }}0.1 + 0.4\)

\({\text{P}}(B) = 0.5\)    A1     N2

[3 marks]

This question was well done by most candidates. In part (b), the intersection \({\text{P}}(A \cap B)\) was sometimes overlooked, incorrectly using \({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B)\) instead.

This question was well done by most candidates. In part (b), the intersection \({\text{P}}(A \cap B)\) was sometimes overlooked, incorrectly using \({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B)\) instead.

Find \({\text{P}}(B)\).

Find \({\text{P}}(A \cup B)\).

valid interpretation (may be seen on a Venn diagram)     (M1)

eg\(\,\,\,\,\,\)\({\text{P}}(A \cap B) + {\text{P}}(A' \cap B),{\text{ }}0.2 + 0.6\)

\({\text{P}}(B) = 0.8\)     A1     N2

[2 marks]

valid attempt to find \({\text{P}}(A)\)     (M1)

eg\(\,\,\,\,\,\)\({\text{P}}(A \cap B) = {\text{P}}(A) \times {\text{P}}(B),{\text{ }}0.8 \times A = 0.2\)

correct working for \({\text{P}}(A)\)     (A1)

eg\(\,\,\,\,\,\)\(0.25,{\text{ }}\frac{{0.2}}{{0.8}}\)

correct working for \({\text{P}}(A \cup B)\)     (A1)

eg\(\,\,\,\,\,\)\(0.25 + 0.8 - 0.2,{\text{ }}0.6 + 0.2 + 0.05\)

\({\text{P}}(A \cup B) = 0.85\)     A1     N3

[4 marks]

Find the probability that the dart does not hit the board.

The contestant scores points as shown in the following table.

Given that the game is fair, find the value of \(q\).

evidence of summing probabilities to \(1\)     (M1)

eg\(\;\;\;\frac{5}{{20}} + \frac{4}{{20}} + \frac{1}{{20}} + p = 1,\;\;\;\sum { = 1} \)

correct working     (A1)

eg\(\;\;\;p = 1 - \frac{{10}}{{20}}\)

\(p = \frac{{10}}{{20}}\;\;\;\left( { = \frac{1}{2}} \right)\)     A1     N2

[3 marks]

correct substitution into \({\text{E}}(X)\)     (A1)

eg\(\;\;\;\frac{4}{{20}}(q) + \frac{1}{{20}}(10) + \frac{{10}}{{20}}( - 3)\)

valid reasoning for fair game (seen anywhere, including equation)     (M1)

eg\(\;\;\;{\text{E}}(X) = 0\), points lost\( = \)points gained

correct working     (A1)

eg\(\;\;\;4q + 10 - 30 = 0,\;\;\;\frac{4}{{20}}q + \frac{{10}}{{20}} = \frac{{30}}{{20}}\)

\(q = 5\)     A1     N2

[4 marks]

Total [7 marks]

The large majority of candidates answered part (a) of the question correctly by summing the probabilities to 1.

Part (b), however was not as well done. Many candidates seemed to be unfamiliar with the idea of a "fair game", despite this topic being listed in the syllabus. The most common error in part (b) was setting \(E(X) = 1\) rather than \(E(X) = 0\).

Find the median mark.

Find the interquartile range.

evidence of median position     (M1)

e.g. 50, line on sketch

median is 56     A1     N2

[2 marks]

lower quartile \(= 40\) , upper quartile \(= 70\)     (A1)(A1)

interquartile range \(= 30\)     A1     N3

[3 marks]

Overall, this question was done well by candidates. In part (a), a surprising number of candidates found the median position (the cumulative frequency) on the y-axis, but did not find the median mark on the x-axis.

Overall, this question was done well by candidates. In part (a), a surprising number of candidates found the median position (the cumulative frequency) on the y-axis, but did not find the median mark on the x-axis. Similar misunderstanding was shown by some candidates in part (b), when attempting to find the interquartile range.

Find the value of the interquartile range.

One student sent k text messages, where k > 11 . Given that k is an outlier, find the least value of k.

recognizing Q₁ or Q₃ (seen anywhere)     (M1)

eg    4,11 , indicated on diagram

IQR = 7     A1 N2

[2 marks]

recognizing the need to find 1.5 IQR     (M1)

eg   1.5 × IQR, 1.5 × 7

valid approach to find k     (M1)

eg   10.5 + 11, 1.5 × IQR + Q₃

21.5     (A1)

k = 22     A1 N3

Note: If no working shown, award N2 for an answer of 21.5.

[4 marks]

The random variable X has the following probability distribution.

Given that \({\rm{E}}(X) = 1.7\) , find q .

correct substitution into \({\rm{E}}(X) = \sum {px} \) (seen anywhere)     A1

e.g. \(1s + 2 \times 0.3 + 3q = 1.7\) , \(s + 3q = 1.1\)

recognizing \(\sum {p = 1} \) (seen anywhere)     (M1)

correct substitution into \(\sum {p = 1} \)     A1

e.g. \(s + 0.3 + q = 1\)

attempt to solve simultaneous equations     (M1)

correct working     (A1)

e.g. \(0.3 + 2q = 0.7\) , \(2s = 1\)

\(q = 0.2\)     A1     N4

[6 marks]

Candidates generally earned either full marks or only one mark on this question. The most common error was where candidates only wrote the equation for \({\rm{E}}(X) = 1.7\) , and tried to rearrange that equation to solve for q. The candidates who also knew that the sum of the probabilities must be equal to 1 were very successful in solving the resulting system of equations.

(i)     Write down the value of \(p\).

(ii)     Find the value of \(q\).

(iii)     Write down the value of \(r\) and of \(s\).

(i)     Write down the probability that this student owns a laptop.

(ii)     Find the probability that this student owns a laptop or a tablet but not both.

(i)     Copy and complete the following tree diagram. (Do not write on this page.)

(ii)     Write down the probability that the second student owns a laptop given that the first owns a laptop.

(i)     \(p = 3\)     A1     N1

(ii)     valid approach     (M1)

eg\(\,\,\,\,\,\)\((12 + 10 + 3) - 21,{\text{ }}22 - 18\)

\(q = 4\)    A1     N2

(iii)     \(r = 8,{\text{ }}s = 6\)     A1A1     N2

(i)     \(\frac{{12}}{{21}}{\text{ }}\left( { = \frac{4}{7}} \right)\)     A2     N2

(ii)     valid approach     (M1)

eg\(\,\,\,\,\,\)\(8 + 6,{\text{ }}r + s\)

\(\frac{{14}}{{21}}{\text{ }}\left( { = \frac{2}{3}} \right)\)    A1     N2

(i)          A1A1A1     N3

(ii)     \(\frac{{11}}{{20}}\)     A1     N1

[4 marks]

On the whole, candidates were very successful on this question, with the majority of candidates earning most of the available marks.

On the whole, candidates were very successful on this question, with the majority of candidates earning most of the available marks.

On the whole, candidates were very successful on this question, with the majority of candidates earning most of the available marks. The most common error was seen in part (c)(ii), where many candidates did not earn the mark. It is also interesting to note that many of the candidates who answered this part correctly did so by using the formula for conditional probability, rather than recognizing that the required probability is given to them in the second branch of the tree diagram.

Copy and complete the table below to show all nine equally likely outcomes.

Let S be the sum of the numbers on the two cards.

Find the probability of each value of S.

Find the expected value of S.

Anna plays a game where she wins \(\$ 50\) if S is even and loses \(\$ 30\) if S is odd.

Anna plays the game 36 times. Find the amount she expects to have at the end of the 36 games.

     A2     N2

[2 marks]

\({\rm{P}}(12) = \frac{1}{9}\) , \({\rm{P}}(13) = \frac{3}{9}\) , \({\rm{P}}(14) = \frac{3}{9}\) , \({\rm{P}}(15) = \frac{2}{9}\)     A2     N2

[2 marks]

correct substitution into formula for \({\text{E}}(X)\)     A1

e.g. \({\rm{E}}(S) = 12 \times \frac{1}{9} + 13 \times \frac{3}{9} + 14 \times \frac{3}{9} + 15 \times \frac{2}{9}\)

\({\rm{E}}(S) = \frac{{123}}{9}\)     A2     N2

[3 marks]

METHOD 1

correct expression for expected gain E(A) for 1 game     (A1)

e.g. \(\frac{4}{9} \times 50 - \frac{5}{9} \times 30\)

\({\rm{E}}(A) = \frac{{50}}{9}\)

amount at end = expected gain for 1 game \( \times 36\)     (M1)

= 200 (dollars)     A1     N2

METHOD 2

attempt to find expected number of wins and losses     (M1)

e.g. \(\frac{4}{9} \times 36\) , \(\frac{5}{9} \times 36\)

attempt to find expected gain E(G)     (M1)

e.g. \(16 \times 50 - 30 \times 20\)

\({\text{E}}(G) = 200\) (dollars)     A1     N2

[3 marks]

Most candidates completed parts (a), (b) and (c) successfully.

Most candidates completed part (b) successfully.

Many found the expected value correctly, while some showed difficulty with the arithmetic.

This was often left blank or only superficially attempted. Some found the expected value \(\frac{{50}}{9}\) but did not answer the question about the amount of money.

Write down the independent variable.

Write down the boiling temperature of the liquid.

Jim describes the correlation as very strong. Circle the value below which best represents the correlation coefficient.

\[0.992\quad \quad \quad 0.251\quad \quad \quad 0\quad \quad \quad  - 0.251\quad \quad \quad  - 0.992\]

Jim’s model is \(d =  - 2.24t + 105\), for \(0 \leqslant t \leqslant 20\). Use his model to predict the decrease in temperature for any 2 minute interval.

\(t\)     A1     N1

[1 mark]

105     A1     N1

[1 mark]

\( - 0.992\)     A2     N2

[2 marks]

valid approach     (M1)

eg\(\,\,\,\,\,\)\(\frac{{{\text{d}}d}}{{{\text{d}}t}} =  - 2.24;{\text{ }}2 \times 2.24,{\text{ }}2 \times  - 2.24,{\text{ }}d(2) =  - 2 \times 2.24 \times 105,\)

finding \(d({t_2}) - d({t_1})\) where \({t_2} = {t_1} + 2\)

4.48 (degrees)     A1     N2

Notes:     Award no marks for answers that directly use the table to find the decrease in temperature for 2 minutes eg \(\frac{{105 - 98.4}}{2} = 3.3\).

[2 marks]

Copy and complete the following tree diagram. (Do not write on this page.)

Find the probability that Bill wins the first game and Andrea wins the second game.

Find the probability that Bill wins at least one game.

Given that Bill wins at least one game, find the probability that he wins both games.

     A1A1A1     N3

Note:   Award A1 for each correct bold probability.

[3 marks]

multiplying along the branches (may be seen on diagram)     (M1)

eg   \(\frac{4}{5} \times \frac{1}{6}\)

\(\frac{4}{{30}}\left( {\frac{2}{{15}}} \right)\)     A1     N2

[2 marks]

METHOD 1

multiplying along the branches (may be seen on diagram)     (M1)

eg   \(\frac{4}{5} \times \frac{5}{6},{\text{ }}\frac{4}{5} \times \frac{1}{6},{\text{ }}\frac{1}{5} \times \frac{2}{3}\)

adding their probabilities of three mutually exclusive paths     (M1)

eg   \(\frac{4}{5} \times \frac{5}{6} + \frac{4}{5} \times \frac{1}{6} + \frac{1}{5} \times \frac{2}{3},{\text{ }}\frac{4}{5} + \frac{1}{5} \times \frac{2}{3}\)

correct simplification     (A1)

eg   \(\frac{{20}}{{30}} + \frac{4}{{30}} + \frac{2}{{15}},{\text{ }}\frac{2}{3} + \frac{2}{{15}} + \frac{2}{{15}}\)

\(\frac{{28}}{{30}}{\text{ }}\left( { = \frac{{14}}{{15}}} \right)\)     A1     N3

METHOD 2

recognizing “Bill wins at least one” is complement of “Andrea wins 2”     (R1)

eg   finding P (Andrea wins 2)

P (Andrea wins both) \( = \frac{1}{5} \times \frac{1}{3}\)     (A1)

evidence of complement     (M1)

eg   \(1 - p,{\text{ }}1 - \frac{1}{{15}}\)

\(\frac{{14}}{{15}}\)     A1     N3

[4 marks]

P (B wins both) \(\frac{4}{5} \times \frac{5}{6}{\text{ }}\left( { = \frac{2}{3}} \right)\)     A1

evidence of recognizing conditional probability     (R1)

eg   \({\text{P}}(A\left| B \right.),{\text{ P (Bill wins both}}\left| {{\text{Bill wins at least one), tree diagram}}} \right.\)

correct substitution     (A2)

eg   \(\frac{{\frac{4}{5} \times \frac{5}{6}}}{{\frac{{14}}{{15}}}}\)

\(\frac{{20}}{{28}}{\text{ }}\left( { = \frac{5}{7}} \right)\)     A1     N3

[5 marks]

(i)     Find \({\rm{P}}(X = 6)\) .

(ii)    Find \({\rm{P}}(X > 6)\) .

(iii)   Find \({\rm{P}}(X = 7|X > 6)\) .

Elena plays a game where she tosses two dice.

If the sum is 6, she wins 3 points.

If the sum is greater than 6, she wins 1 point.

If the sum is less than 6, she loses k points.

Find the value of k for which the game is fair.

(i) number of ways of getting \(X = 6\) is 5     A1

\({\rm{P}}(X = 6) = \frac{5}{{36}}\)     A1     N2

(ii) number of ways of getting \(X > 6\) is 21     A1

\({\rm{P}}(X > 6) = \frac{{21}}{{36}}\left( { = \frac{7}{{12}}} \right)\)     A1     N2

(iii) \({\rm{P}}(X = 7|X > 6) = \frac{6}{{21}}\left( { = \frac{2}{7}} \right)\)     A2     N2

[6 marks]

attempt to find \({\rm{P}}(X < 6)\)     M1

e.g. \(1 - \frac{5}{{36}} - \frac{{21}}{{36}}\)

\({\rm{P}}(X < 6) = \frac{{10}}{{36}}\)     A1

fair game if \({\rm{E}}(W) = 0\) (may be seen anywhere)     R1

attempt to substitute into \({\rm{E}}(X)\) formula     M1

e.g. \(3\left( {\frac{5}{{36}}} \right) + 1\left( {\frac{{21}}{{36}}} \right) - k\left( {\frac{{10}}{{36}}} \right)\)

correct substitution into \({\rm{E}}(W) = 0\)     A1

e.g. \(3\left( {\frac{5}{{36}}} \right) + 1\left( {\frac{{21}}{{36}}} \right) - k\left( {\frac{{10}}{{36}}} \right) = 0\)

work towards solving     M1

e.g. \(15 + 21 - 10k = 0\)

\(36 = 10k\)     A1

\(k = \frac{{36}}{{10}}( = 3.6)\)     A1     N4

[8 marks]

Write down the possible minimum and maximum values of r .

Given that \(r = 0.95\) , which of the following diagrams best represents the data.

For the data in diagram D , which two of the following expressions describe the correlation between x and y ?

perfect, zero, linear, strong positive, strong negative, weak positive, weak negative

min value of r is \( - 1\), max value of r is 1     A1A1    N2

[2 marks]

C     A1     N1

[1 mark]

linear, strong negative     A1A1     N2

[2 marks]

Calculate the value of k.

Find

(i)     the median;

(ii)    the interquartile range.

evidence of using \(\sum {{f_i} = 100} \)     (M1)

\(k = 4\)     A1     N2

[2 marks]

(i) evidence of median position     (M1)

e.g. 50th item, \(26 + 10 + 20 = 56\)

\({\text{median}} = 3\)     A1     N2

(ii) \({Q_1} = 1\) and \({Q_3} = 5\)     (A1)(A1)

\({\text{interquartile range}} = 4\) (accept 1 to 5 or \(5 - 1\) , etc.)     A1     N3

[5 marks]

Frequencies and median seemed well understood, but quartiles and inter-quartile range less so.

Frequencies and median seemed well-understood, but quartiles and interquartile range less so. A few students, probably based on past papers, drew cumulative frequency diagrams, generating slightly different answers for median and quartiles.

Find the median number of hours worked by the employees.

Write down the number of employees who worked 50 hours or less.

Find the amount of money an employee earned for working 40 hours;

Find the amount of money an employee earned for working 43 hours.

Find the number of employees who earned £200 or less.

Only 10 employees earned more than £\(k\). Find the value of \(k\).

evidence of median position     (M1)

eg\(\,\,\,\,\,\)80th employee

40 hours     A1     N2

[2 marks]

130 employees     A1     N1

[1 mark]

£320     A1     N1

[1 mark]

splitting into 40 and 3     (M1)

eg\(\,\,\,\,\,\)3 hours more, \(3 \times 10\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(320 + 3 \times 10\)

£350     A1     N3

[3 marks]

valid approach     (M1)

eg\(\,\,\,\,\,\)200 is less than 320 so 8 pounds/hour, \(200 \div 8,{\text{ }}25,{\text{ }}\frac{{200}}{{320}} = \frac{x}{{40}}\),

18 employees     A2     N3

[3 marks]

valid approach     (M1)

eg\(\,\,\,\,\,\)\(160 - 10\)

60 hours worked     (A1)

correct working     (A1)

eg\(\,\,\,\,\,\)\(40(8) + 20(10),{\text{ }}320 + 200\)

\(k = 520\)     A1     N3

[4 marks]

Write down the value of t .

(i)     Show that \(r = 0.2\) .

(ii)    Write down the value of q and of s .

(i)     Write down \({\rm{P}}(B')\) .

(ii)    Find \({\rm{P}}(A|B')\) .

\(t = 0.3\)     A1     N1

[1 mark]

(i) correct values     A1

e.g. \(0.3 + 0.6 - 0.7\) , \(0.9 - 0.7\)

\(r = 0.2\)     AG     N0

(ii) \(q = 0.1\) , \(s = 0.4\)     A1A1     N2

[3 marks]

(i) \(0.4\)     A1     N1

(ii) \({\rm{P}}(A|B') = \frac{1}{4}\)     A2     N2

[3 marks]

Parts (a), (b), and (c)(i) of this Venn diagram probability question were answered quite well with candidates consistently earning full marks.

Parts (a), (b), and (c)(i) of this Venn diagram probability question were answered quite well with candidates consistently earning full marks. Only a few candidates worked backwards from the \(r = 0.2\) given in the "show that" portion of part (b).

Many candidates struggled on part (c)(ii), either not recognizing conditional probability or multiplying probabilities to find the numerator as if the events were independent. A number of candidates who successfully found the probability in part (c)(ii) left their incomplete answer of \(\frac{{0.1}}{{0.4}}\) .

(i)     Find the number of boys who play both sports.

(ii)    Write down the number of boys who play only rugby.

One boy is selected at random.

(i)     Find the probability that he plays only one sport.

(ii)    Given that the boy selected plays only one sport, find the probability that he plays rugby.

Let A be the event that a boy plays football and B be the event that a boy plays rugby.

Explain why A and B are not mutually exclusive.

Show that A and B are not independent.

(i) evidence of substituting into \(n(A \cup B) = n(A) + n(B) - n(A \cap B)\)     (M1)

e.g. \(75 + 55 - 100\) , Venn diagram

30     A1     N2 

(ii) 45     A1     N1

[3 marks]

(i) METHOD 1

evidence of using complement, Venn diagram     (M1)

e.g. \(1 - p\) , \(100 - 30\)

\(\frac{{70}}{{100}}\) \(\left( { = \frac{7}{{10}}} \right)\)     A1     N2

METHOD 2

attempt to find P(only one sport) , Venn diagram     (M1) 

e.g. \(\frac{{25}}{{100}} + \frac{{45}}{{100}}\)

\(\frac{{70}}{{100}}\) \(\left( { = \frac{7}{{10}}} \right)\)     A1     N2

(ii) \(\frac{{45}}{{70}}\) \(\left( { = \frac{9}{{14}}} \right)\)     A2     N2

[4 marks]  

valid reason in words or symbols     (R1)

e.g. \({\rm{P}}(A \cap B) = 0\) if mutually exclusive, \({\rm{P}}(A \cap B) \ne 0\) if not mutually exclusive

correct statement in words or symbols     A1     N2

e.g. \({\rm{P}}(A \cap B) = 0.3\) , \({\rm{P}}(A \cup B) \ne {\rm{P}}(A) + {\rm{P}}(B)\) , \({\rm{P}}(A) + {\rm{P}}(B) > 1\) , some students play both sports, sets intersect

[2 marks]

valid reason for independence     (R1)

e.g. \({\rm{P}}(A \cap B) = {\rm{P}}(A) \times {\rm{P}}(B)\) , \({\rm{P}}(B|A) = {\rm{P}}(B)\)

correct substitution     A1A1     N3

e.g. \(\frac{{30}}{{100}} \ne \frac{{75}}{{100}} \times \frac{{55}}{{100}}\) , \(\frac{{30}}{{55}} \ne \frac{{75}}{{100}}\)

[3 marks]

Overall, this question was very well done. There were some problems with the calculation of conditional probability, where a considerable amount of candidates tried to use a formula instead of using its concept and analysing the problem. It is the kind of question where it can be seen if the concept is not clear to candidates.

Overall, this question was very well done. There were some problems with the calculation of conditional probability, where a considerable amount of candidates tried to use a formula instead of using its concept and analysing the problem. It is the kind of question where it can be seen if the concept is not clear to candidates.

In part (c), candidates were generally able to explain in words why events were mutually exclusive, though many gave the wrong values for P(A) and P(B).

There was a great amount of confusion between the concepts of independent and mutually exclusive events. In part (d), the explanations often referred to mutually exclusive events.

It was evident that candidates need more practice with questions like (c) and (d).

Some students equated probabilities and number of elements, giving probabilities greater than 1.

Write down the value of p .

Find \({\rm{P}}(B)\) .

Find \({\rm{P}}(A'|B)\) .

\(p = \frac{4}{5}\)     A1     N1

[1 mark]

multiplying along the branches     (M1)

e.g. \(\frac{1}{5} \times \frac{1}{4}\) , \(\frac{{12}}{{40}}\)

adding products of probabilities of two mutually exclusive paths     (M1)

e.g. \(\frac{1}{5} \times \frac{1}{4} + \frac{4}{5} \times \frac{3}{8}\) , \(\frac{1}{{20}} + \frac{{12}}{{40}}\)

\({\rm{P}}(B) = \frac{{14}}{{40}}\) \(\left( { = \frac{7}{{20}}} \right)\)     A1     N2

[3 marks]

appropriate approach which must include \({A'}\) (may be seen on diagram)     (M1)

e.g. \(\frac{{{\rm{P}}(A' \cap B)}}{{{\rm{P}}(B)}}\) (do not accept \(\frac{{{\rm{P}}(A \cap B)}}{{{\rm{P}}(B)}}\) )

\({\rm{P}}(A'|B) = \frac{{\frac{4}{5} \times \frac{3}{8}}}{{\frac{7}{{20}}}}\)     (A1)

\({\rm{P}}(A'|B) = \frac{{12}}{{14}}\) \(\left( { = \frac{6}{7}} \right)\)     A1     N2

[3 marks]

While nearly every candidate answered part (a) correctly, many had trouble with the other parts of this question.

In part (b), many candidates did not multiply along the branches of the tree diagram to find the required values, and many did not realize that there were two paths for P(B). There were also many candidates who understood what the question required, but then did not know how to multiply fractions correctly, and these calculation errors led to an incorrect answer.

In part (c), most candidates attempted to use a formula for conditional probability found in the information booklet, but very few substituted the correct values.

Celeste wishes to hire a taxicab from a company which has a large number of taxicabs.

The taxicabs are randomly assigned by the company.

     The probability that a taxicab is yellow is 0.4.

     The probability that a taxicab is a Fiat is 0.3.

     The probability that a taxicab is yellow or a Fiat is 0.6.

 Find the probability that the taxicab hired by Celeste is not a yellow Fiat.

recognize need for intersection of Y and F     (R1)

eg     \({\text{P}}(Y \cap F){\text{, }}0.3 \times 0.4\)

valid approach to find \({\text{P}}(Y \cap F)\)     (M1)

eg     \({\text{P}}(Y) + {\text{P}}(F) - {\text{P}}(Y \cup F)\), Venn diagram

correct working (may be seen in Venn diagram)     (A1)

eg     \(0.4 + 0.3 - 0.6\)

\({\text{P}}(Y \cap F) = 0.1\)     A1

recognize need for complement of \(Y \cap F\)     (M1)

eg     \(1 - {\text{P}}(Y \cap F){\text{, }}1 - 0.1\)

\({\text{P}}\left( {(Y \cap F)'} \right) = 0.9\)     A1     N3

[6 marks]

Show that \(\cos \theta  = \frac{3}{4}\).

Given that \(\tan \theta  > 0\), find \(\tan \theta \).

Let \(y = \frac{1}{{\cos x}}\), for \(0 < x < \frac{\pi }{2}\). The graph of \(y\)between \(x = \theta \) and \(x = \frac{\pi }{4}\) is rotated 360° about the \(x\)-axis. Find the volume of the solid formed.

evidence of summing to 1     (M1)

eg\(\,\,\,\,\,\)\(\sum {p = 1} \)

correct equation     A1

eg\(\,\,\,\,\,\)\(\cos \theta  + 2\cos 2\theta  = 1\)

correct equation in \(\cos \theta \)     A1

eg\(\,\,\,\,\,\)\(\cos \theta  + 2(2{\cos ^2}\theta  - 1) = 1,{\text{ }}4{\cos ^2}\theta  + \cos \theta  - 3 = 0\)

evidence of valid approach to solve quadratic     (M1)

eg\(\,\,\,\,\,\)factorizing equation set equal to \(0,{\text{ }}\frac{{ - 1 \pm \sqrt {1 - 4 \times 4 \times ( - 3)} }}{8}\)

correct working, clearly leading to required answer     A1

eg\(\,\,\,\,\,\)\((4\cos \theta  - 3)(\cos \theta  + 1),{\text{ }}\frac{{ - 1 \pm 7}}{8}\)

correct reason for rejecting \(\cos \theta  \ne  - 1\)     R1

eg\(\,\,\,\,\,\)\(\cos \theta \) is a probability (value must lie between 0 and 1), \(\cos \theta  > 0\)

Note:     Award R0 for \(\cos \theta  \ne  - 1\) without a reason.

\(\cos \theta  = \frac{3}{4}\)    AG  N0

valid approach     (M1)

eg\(\,\,\,\,\,\)sketch of right triangle with sides 3 and 4, \({\sin ^2}x + {\cos ^2}x = 1\)

correct working     

(A1)

eg\(\,\,\,\,\,\)missing side \( = \sqrt 7 ,{\text{ }}\frac{{\frac{{\sqrt 7 }}{4}}}{{\frac{3}{4}}}\)

\(\tan \theta  = \frac{{\sqrt 7 }}{3}\)     A1     N2

[3 marks]

attempt to substitute either limits or the function into formula involving \({f^2}\)     (M1)

eg\(\,\,\,\,\,\)\(\pi \int_\theta ^{\frac{\pi }{4}} {{f^2},{\text{ }}\int {{{\left( {\frac{1}{{\cos x}}} \right)}^2}} } \)

correct substitution of both limits and function     (A1)

eg\(\,\,\,\,\,\)\(\pi \int_\theta ^{\frac{\pi }{4}} {{{\left( {\frac{1}{{\cos x}}} \right)}^2}{\text{d}}x} \)

correct integration     (A1)

eg\(\,\,\,\,\,\)\(\tan x\)

substituting their limits into their integrated function and subtracting     (M1)

eg\(\,\,\,\,\,\)\(\tan \frac{\pi }{4} - \tan \theta \)

Note:     Award M0 if they substitute into original or differentiated function.

\(\tan \frac{\pi }{4} = 1\)    (A1)

eg\(\,\,\,\,\,\)\(1 - \tan \theta \)

\(V = \pi  - \frac{{\pi \sqrt 7 }}{3}\)     A1     N3

[6 marks]

Find the range of the lengths of all 200 fish.

Four cumulative frequency graphs are shown below.

Which graph is the best representation of the lengths of the female fish?

correct end points     (A1)(A1)

max = 27 , min = 4

range = 23     A1     N3

[3 marks]

Graph 3     A2     N2

[2 marks]

While there were a large number of candidates who answered both parts of this question correctly, a surprising number did not know how to find the range of all 200 fish in part (a). Common errors included finding the ranges of the male and female fish separately, or averaging the separate ranges of the male and female fish.

Some candidates did not interpret the cumulative frequency graphs correctly, or just seemed to guess which graph was correct. The most common incorrect "guess" was graph 4, likely because this graph had a more familiar cumulative shape.

Complete the following tree diagram.

Find the probability that exactly one of the selected balls is green.

correct probabilities

     A1A1A1     N3

Note:     Award A1 for each correct bold answer.

[3 marks]

multiplying along branches     (M1)

eg\(\,\,\,\,\,\)\(\frac{5}{8} \times \frac{3}{7},{\text{ }}\frac{3}{8} \times \frac{5}{7},{\text{ }}\frac{{15}}{{56}}\)

adding probabilities of correct mutually exclusive paths     (A1)

eg\(\,\,\,\,\,\)\(\frac{5}{8} \times \frac{3}{7} + \frac{3}{8} \times \frac{5}{7},{\text{ }}\frac{{15}}{{56}} + \frac{{15}}{{56}}\)

\(\frac{{30}}{{56}}{\text{ }}\left( { = \frac{{15}}{{28}}} \right)\)     A1     N2

[3 marks]

Find the value of r .

Given that \({\rm{E}}(X) = 1.4\) , find the value of p and of q .

attempt to substitute \({\rm{P}}(X > 1) = 0.5\)     (M1)

e.g. \(r + 0.2 = 0.5\)

\(r = 0.3\)     A1     N2

[2 marks]

correct substitution into \({\rm{E}}(X)\) (seen anywhere)     (A1)

e.g. \(0 \times p + 1 \times q + 2 \times r + 3 \times 0.2\)

correct equation     A1

e.g. \(q + 2 \times 0.3 + 3 \times 0.2 = 1.4\) , \(q + 1.2 = 1.4\)

\(q = 0.2\)     A1      N1

evidence of choosing \(\sum {{p_i} = 1} \)     M1

e.g. \(p + 0.2 + 0.3 + 0.2 = 1\) , \(p + q = 0.5\)

correct working     (A1)

\(p + 0.7 = 1\) , \(1 - 0.2 - 0.3 - 0.2\) , \(p + 0.2 = 0.5\)

\(p = 0.3\)     A1 N2

Note: Exception to the FT rule. Award FT marks on an incorrect value of q, even if q is an inappropriate value. Do not award the final A mark for an inappropriate value of p.

[6 marks]

The majority of candidates were successful in earning full marks on this question.

In part (b), a small number of candidates did not use the correct formula for \({\rm{E}}(X)\) , even though this formula is given in the formula booklet. There were also a few candidates who incorrectly assumed that \(p = 0\) , forgetting that the sum of the probabilities must equal 1. There were a few candidates who left this question blank, which raises concerns about whether they had been exposed to probability distributions during the course.

Find the median number of cans collected.

(i)     Write down the value of \(a\).

(ii)     The interquartile range is 14. Find the value of \(b\).

Sam’s class collected 745 cans. They want an average of 40 cans per student.

How many more cans need to be collected to achieve this target?

The students raise $0.10 for each recycled can.

(i)     Find the largest amount raised by a student in Sam’s class.

(ii)     The following cumulative frequency curve shows the amounts in dollars raised by all the students in the school. Find the percentage of students in the school who raised more money than anyone in Sam’s class.

(i)     Write down the new mean.

(ii)     Write down the new standard deviation.

valid approach     (M1)

eg\(\,\,\,\,\,\)between 10th and 11th, \(\frac{{8 + 8}}{2}\)

median \( = 38\)     A1     N2

[2 marks]

(i)     \(a = 20\)     A1     N1

(ii)     valid approach     (M1)

eg\(\,\,\,\,\,\)\({Q_3} - {Q_1},{\text{ }}{Q_1} + 14,{\text{ }}b - 30 = 14\)

\(b = 44\)    A1     N2

[3 marks]

valid approach     (M1)

egx\(\,\,\,\,\,\)\(40 \times 20,{\text{ }}\frac{{x + 745}}{{20}},{\text{ }}40 - \frac{{745}}{{20}}\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(800 - 745,{\text{ }}20 \times 2.75\)

55 (more cans)     A1     N2

[3 marks]

(i)     most cans in Sam’s class \( = 50\)     (A1)

5 ($)     A1     N2

(ii)     correct value of 64 or 16     A1

valid approach     (M1)

eg\(\,\,\,\,\,\)\(\frac{{64}}{{80}},{\text{ }}80\% ,{\text{ }}80 - 64,{\text{ }}\frac{{16}}{{80}}\)

20%     A1     N2

[5 marks]

(i)     41.4 (exact)     A1 N1

(ii)     18.5     A1     N1

[2 marks]

Generally, candidates were very successful with this question, appearing to move easily between the three different representations of data. The main conceptual errors appeared in part d) where a percentage of 100 was found (instead of 80) and in part e) where the new standard deviation was often given as 20.5. Arithmetic errors seemed to be the other factor, with a surprising number of candidates finding in part c) that \(800 - 745 = 15\).

Generally, candidates were very successful with this question, appearing to move easily between the three different representations of data. The main conceptual errors appeared in part d) where a percentage of 100 was found (instead of 80) and in part e) where the new standard deviation was often given as 20.5. Arithmetic errors seemed to be the other factor, with a surprising number of candidates finding in part c) that \(800 - 745 = 15\).

Generally, candidates were very successful with this question, appearing to move easily between the three different representations of data. The main conceptual errors appeared in part d) where a percentage of 100 was found (instead of 80) and in part e) where the new standard deviation was often given as 20.5. Arithmetic errors seemed to be the other factor, with a surprising number of candidates finding in part c) that \(800 - 745 = 15\).

Generally, candidates were very successful with this question, appearing to move easily between the three different representations of data. The main conceptual errors appeared in part d) where a percentage of 100 was found (instead of 80) and in part e) where the new standard deviation was often given as 20.5. Arithmetic errors seemed to be the other factor, with a surprising number of candidates finding in part c) that \(800 - 745 = 15\).

Generally, candidates were very successful with this question, appearing to move easily between the three different representations of data. The main conceptual errors appeared in part d) where a percentage of 100 was found (instead of 80) and in part e) where the new standard deviation was often given as 20.5. Arithmetic errors seemed to be the other factor, with a surprising number of candidates finding in part c) that \(800 - 745 = 15\).

Copy and complete the following tree diagram.

Find the probability that Pablo leaves home before 07:00 and is late for work.

Find the probability that Pablo is late for work.

Given that Pablo is late for work, find the probability that he left home before 07:00.

Two days next week Pablo will drive to work. Find the probability that he will be late at least once.

A1A1A1 N3

Note: Award A1 for each bold fraction.

[3 marks]

multiplying along correct branches      (A1)
eg  \(\frac{3}{4} \times \frac{1}{8}\)

P(leaves before 07:00 ∩ late) = \(\frac{3}{32}\)    A1 N2

[2 marks]

multiplying along other “late” branch      (M1)
eg  \(\frac{1}{4} \times \frac{5}{8}\)

adding probabilities of two mutually exclusive late paths      (A1)
eg  \(\left( {\frac{3}{4} \times \frac{1}{8}} \right) + \left( {\frac{1}{4} \times \frac{5}{8}} \right),\,\,\frac{3}{{32}} + \frac{5}{{32}}\)

\({\text{P}}\left( L \right) = \frac{8}{{32}}\,\,\left( { = \frac{1}{4}} \right)\)    A1 N2

[3 marks]

recognizing conditional probability (seen anywhere)      (M1)
eg  \({\text{P}}\left( {A|B} \right),\,\,{\text{P}}\left( {{\text{before 7}}|{\text{late}}} \right)\)

correct substitution of their values into formula      (A1)
eg \(\frac{{\frac{3}{{32}}}}{{\frac{1}{4}}}\)

\({\text{P}}\left( {{\text{left before 07:00}}|{\text{late}}} \right) = \frac{3}{8}\)    A1 N2

[3 marks]

valid approach      (M1)
eg  1 − P(not late twice), P(late once) + P(late twice)

correct working      (A1)
eg  \(1 - \left( {\frac{3}{4} \times \frac{3}{4}} \right),\,\,2 \times \frac{1}{4} \times \frac{3}{4} + \frac{1}{4} \times \frac{1}{4}\)

\(\frac{7}{{16}}\)    A1 N2

[3 marks]

(i)     Write down the median weekly wage.

(ii)    Find the interquartile range of the weekly wages.

The box-and-whisker plot below displays the weekly wages of the employees.

Write down the value of

(i)     \(a\) ;

(ii)    \(b\) ;

(iii)   \(c\) .

Employees are paid \($\ 20\) per hour.

Find the median number of hours worked per week.

Employees are paid \(\$ 20\) per hour.

Find the number of employees who work more than \(25\) hours per week.

(i) median weekly wage \(= 400\) (dollars)      A1     N1

(ii) lower quartile \(= 330\), upper quartile \(= 470\)    (A1)(A1)

\({\text{IQR}} = 140\) (dollars) (accept any notation suggesting interval \(330\) to \(470\))    A1     N3

Note: Exception to the FT rule. Award A1(FT) for an incorrect IQR only if both quartiles are explicitly noted.

[4 marks]

(i) \(330\) (dollars)     A1     N1

(ii) \(400\) (dollars)    A1     N1

(iii) \(700\) (dollars)   A1     N1

[3 marks]

valid approach     (M1)

e.g. \({\rm{hours = }}\frac{{{\rm{wages}}}}{{{\rm{rate}}}}\)

correct substitution     (A1)

e.g. \(\frac{{400}}{{20}}\)

median hours per week \(= 20\)     A1     N2

[3 marks]

attempt to find wages for 25 hours per week     (M1)

e.g. \({\text{wages}} = {\text{hours}} \times {\text{rate}}\)

correct substitution     (A1)

e.g. \(25 \times 20\)

finding wages \(= 500\)     (A1)

65 people (earn 500\( \leqslant \))     (A1)

15 people (work more than 25 hours)     A1     N3

[5 marks]

Many candidates answered this question completely correctly, earning full marks in all parts of the question. In parts (a) and (b), there were some who gave the frequency values on the y-axis, rather than the wages on the x-axis, as their quartiles and interquartile range.

Many candidates answered this question completely correctly, earning full marks in all parts of the question. In parts (a) and (b), there were some who gave the frequency values on the y-axis, rather than the wages on the x-axis, as their quartiles and interquartile range.

For part (c), the majority of candidates seemed to understand what was required, though there were a few who used an extreme value such as \(700\), rather than the median value.

In part (d), some candidates simply answered \(65\), which is the number of workers earning \(\$ 500\) or less, rather than finding the number of workers who earned more than \(\$ 500\). It was interesting to note that quite a few candidates gave their final answer as \(14\), rather than \(15\).