2011 H2 Mathematics Paper 2 Question 9

Probability

Answers

{0.058}

{\frac{15}{29}\approx0.517}

{0.109}

{0.313}

(ia)

\begin{align*} & \textrm{P}\left(\textrm{faulty}\right) \\ &= 0.6(0.05) + 0.4(0.07) \\ &= 0.058 \; \blacksquare \end{align*}

(ib)

\begin{align*} &\textrm{P}\left(A \mid \textrm{faulty}\right) \\ &=\frac{\textrm{P}\left(A \cap \textrm{faulty}\right)}{\textrm{P}\left(\textrm{faulty}\right)} \\ &= \frac{\frac{2}{5}(\frac{7}{100})}{\frac{29}{500}} \\ &= \frac{15}{29} \; \blacksquare \end{align*}

(iia)

\begin{align*} & \textrm{P}\left(\textrm{exactly one faulty}\right) \\ & = \textrm{P}\left(\textrm{faulty}\right) \times \textrm{P}\left(\textrm{not faulty}\right) \times 2! \\ & = 0.058 \times (1-0.058) \times 2! \\ & = 0.109 \; (3\textrm{s.f.}) \; \blacksquare \end{align*}

(iib)

\begin{align*} &\textrm{P}\left(\textrm{both by }A \mid \textrm{exactly one faulty}\right) \\ &=\frac{\textrm{P}\left(\textrm{both by }A \cap \textrm{exactly one faulty}\right)}{\textrm{P}\left(\textrm{exactly one faulty}\right)} \\ &= \frac{\textrm{P}\left(A,\textrm{ faulty}\right)\times\textrm{P}\left(A,\textrm{ not faulty}\right)\times 2!}{0.10927} \\ &= 0.6(0.05) \times 0.6(1-0.05) \times 2! \\ &= 0.313 \; (3\textrm{s.f.}) \; \blacksquare \end{align*}