Quote:

Originally Posted by **mika**

There are two diagnostic tests for a disease. Among those who have the disease, 10% give negative results for the first test and, independently of this, 5% give negative results on the second test. Among those who do not have the disease, 80% give negative results on the first test and, independently, 70% give negative results on the second test. From historical data it is known that 20% of those sent by their GPs for diagnostic testing actually have the disease.

c) If the first test gives a positive result, what is the probability that the second test will also be positive?

If the first test gives a positive result:

$\displaystyle

p(\mbox{infected}|\mbox{pos on t1})=\frac{p(\mbox{pos t1}|\mbox{infected})p_0}{p(\mbox{pos t1})}

$

$\displaystyle

=\frac{0.9 \times 0.2}{0.9 \times 0.2+0.2 \times 0.8}\approx 0.529

$

$\displaystyle

p(\mbox{uninfctd}|\mbox{pos on t1})=\frac{p(\mbox{positive on t1}|\mbox{uninfctd})(1-p_0)}{p(\mbox{pos on t1})}

$

$\displaystyle

=\frac{0.2 \times 0.8}{0.9 \times 0.2+0.2 \times 0.8}\approx 0.471

$

.

Finally:

$\displaystyle

p(\mbox{pos on t2}|\mbox{pos on t1})=$$\displaystyle p(\mbox{uninfctd}|\mbox{pos on t1})p(\mbox{pos on t2}|\mbox{uninfctd})$$\displaystyle +p(\mbox{infected}|\mbox{pos on t1})p(\mbox{pos on t2}|\mbox{infected})

$

$\displaystyle

=0.471 \times 0.3+0.529 \times 0.95\approx 0.644

$

RonL