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There are 10 men and 20 women in a locality. Among them half of men and half of women are brown eyed. A person is selected at random from the locality and found to be brown eyed. The probability that the selected person is a man is k then 6k =

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answer is 2.

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Detailed Solution

Let E denotes the event of the person being brown eyed.
$P (M) = \frac{10}{30} = \frac{1}{3}, P (W) = \frac{20}{30} = \frac{2}{3}$
$P (\frac{E}{M}) = \frac{5}{10} = \frac{1}{2}$ , $P (\frac{E}{W}) = \frac{10}{20} = \frac{1}{2}$

$∴ P (\frac{M}{E}) = \frac{P (\frac{E}{M}) . P (M)}{P (\frac{E}{M}) . P (M) + P (\frac{E}{W}) . P (W)}$ ( bayers theorem)
$\frac{\frac{1}{2} . \frac{1}{3}}{\frac{1}{2} . \frac{1}{3} + \frac{1}{2} . \frac{2}{3}} = \frac{1}{3}$

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There are 10 men and 20 women in a locality. Among them half of men and half of women are brown eyed. A person is selected at random from the locality and found to be brown eyed. The probability that the selected person is a man is k then 6k =