H2 Maths A Level 2012 Solution, Paper 2 Q5; H2 Maths Tuition

5(i)(a)

$P(\text{patient has the disease and test positive})=0.001(0.995)=9.95\times 10^{-4}$

$P(\text{patient does not have the disease and he tests positive})=(1-0.001)(1-0.995)=4.995\times 10^{-3}$

$P(\text{result of the test is positive})=9.95\times 10^{-4}+4.995\times 10^{-3}=5.99\times 10^{-3}$

(b)

Let A=patient has disease

Let B=result of test is positive

$\displaystyle\begin{array}{rcl}P(A|B)&=&\frac{P(A\cap B)}{P(B)}\\ &=&\frac{(0.001)(0.995)}{5.99\times 10^{-3}}\\ &=&0.166 \end{array}$

Note that the probability is surprisingly quite low! (This is called the False positive paradox, a statistical result where false positive tests are more probable than true positive tests, occurring when the overall population has a low incidence of a condition and the incidence rate is lower than the false positive rate. See http://en.wikipedia.org/wiki/False_positive_paradox)

(ii)

$\displaystyle P(A|B)=\frac{(0.001)p}{(0.001)p+(1-0.001)(1-p)}=0.75$

By GC, $p=0.999666$ (6 d.p.)

Author: mathtuition88

https://mathtuition88.com/

This site uses Akismet to reduce spam. Learn how your comment data is processed.