2012 H2 Mathematics Paper 2 Question 5

Probability

Answers

(ia)
0.005 99{0.005\,99}
(ib)
0.166{0.166}

(ii)

0.999 666{0.999\,666}

Full solutions

(ia)
P(test positive)=0.001(0.995)+0.999(0.005)=0.005 99  ■\begin{align*} & \textrm{P}\left(\textrm{test positive}\right) \\ &= 0.001(0.995) + 0.999(0.005) \\ &= 0.005\,99 \; \blacksquare \end{align*}
(ib)
P(has disease∣positive)=P(has disease∩positive)P(positive)=0.001(0.995)0.00599=0.166  (3s.f.)  ■\begin{align*} &\textrm{P}\left(\textrm{has disease} \mid \textrm{positive}\right) \\ &=\frac{\textrm{P}\left(\textrm{has disease} \cap \textrm{positive}\right)}{\textrm{P}\left(\textrm{positive}\right)} \\ &= \frac{0.001(0.995)}{0.00599} \\ &= 0.166 \; (3\textrm{s.f.}) \; \blacksquare \end{align*}

(ii)

P(has disease∣positive)=0.75P(has disease∩positive)P(positive)=0.750.001p0.001p+0.999(1−p)=0.75\begin{align*} \textrm{P}\left(\textrm{has disease} \mid \textrm{positive}\right) &= 0.75 \\ \frac{\textrm{P}\left(\textrm{has disease} \cap \textrm{positive}\right)}{\textrm{P}\left(\textrm{positive}\right)} &= 0.75 \\ \frac{0.001p}{0.001p + 0.999(1-p)} &= 0.75 \end{align*}
Hence p=0.999 666  (6d.p.)  ■{p = 0.999\,666 \; (6\textrm{d.p.}) \; \blacksquare}