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# Suppose that 5% of men and 0.25% of women have grey hair. A grey haired person is selected at random. What is the probability of this person being male? Assume that there are equal number of males and females.

• $P \large(\frac{B}{A}) =$$\large (\frac{P(A\;\cap\; B)}{P(A)}) • According to Bayes Theorem, if E_1,E_2,E_3.....E_n are a set of mutually exclusive and exhaustive events, then P\left(\large \frac{E_i}{E}\right ) = \Large \frac{P\left(\frac{E}{E_i}\right ). P(E_i)} {\sum_{i=1}^{n} (P\left(\frac{E}{E_i}\right ).P(E_i))} Let E_1 be the event that the selected person is male and E_2 be the event that its a female. P (E_1) = \large\frac{1}{2} =$$P (E_2)$
$E_1$ and $E_2$ are mutually exclusive and exhaustive, so we can apply Bayes theorem.
Let E: be the event that the person has grey hair. 5% of men picked at random have grey hair $\rightarrow P \large (\frac{E}{E_1}) = \frac{5}{100} = \frac{1}{20}$
We can calcuate the probability: $P \large \left(\frac{E_1}{E}\right) = \Large \frac{P\left(\frac{E}{E_1}\right ). P(E_1)} { (P\left(\frac{E}{E_1}\right ).P(E_1)) + (P\left(\frac{E}{E_2}\right ).P(E_2))}$
$P \large \left(\frac{E_1}{E}\right) = \Large \frac{\frac{1}{20} \frac{1}{2}} { \frac{1}{20}\frac{1}{2} + \frac{1}{400}\frac{1}{2}} = \frac{20}{21}$