What is the probability that there are exactly 3 defective DVD players in the first 8 DVD players examined?

What is the probability that the ${9^{{\text{th}}}}$ DVD player examined is the ${4^{{\text{th}}}}$ defective one found?

METHOD 1

P(3 defective in first 8) $=\left( {\begin{array}{*{20}{c}}   8 \\   3 \end{array}} \right) \times \frac{4}{{15}} \times \frac{3}{{14}} \times \frac{2}{{13}} \times \frac{{11}}{{12}} \times \frac{{10}}{{11}} \times \frac{9}{{10}} \times \frac{8}{9} \times \frac{7}{8}$     M1A1A1

Note: Award M1 for multiplication of probabilities with decreasing denominators.

Award A1 for multiplication of correct eight probabilities.

Award A1 for multiplying by $\left( {\begin{array}{*{20}{c}}   8 \\   3 \end{array}} \right)$.

$= \frac{{56}}{{195}}$     A1 

METHOD 2

P(3 defective DVD players from 8) $= \frac{{\left( {\begin{array}{*{20}{c}}   4 \\   3 \end{array}} \right)\left( {\begin{array}{*{20}{c}}   {11} \\   5 \end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}   {15} \\   8 \end{array}} \right)}}$     M1A1

Note: Award M1 for an expression of this form containing three combinations.

$= \frac{{\frac{{4!}}{{3!1!}} \times \frac{{11!}}{{5!6!}}}}{{\frac{{15!}}{{8!7!}}}}$     M1

$= \frac{{56}}{{195}}$     A1 

[4 marks]

${\text{P(}}{{\text{9}}^{{\text{th}}}}{\text{ selected is }}{{\text{4}}^{{\text{th}}}}{\text{ defective player}}|{\text{3 defective in first 8)}} = \frac{1}{7}$     (A1)

${\text{P(}}{{\text{9}}^{{\text{th}}}}{\text{ selected is }}{{\text{4}}^{{\text{th}}}}{\text{ defective player)}} = \frac{{56}}{{195}} \times \frac{1}{7}$     M1

$= \frac{8}{{195}}$     A1

[3 marks]

There were two main methods used to complete this question, the most common being a combinations approach. Those who did this coped well with the factorial simplification. Many who did not manage the first part were able to complete the second part successfully.

There were two main methods used to complete this question, the most common being a combinations approach. Those who did this coped well with the factorial simplification. Many who did not manage the first part were able to complete the second part successfully.