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In a certain college, 4% of the men and 1% of the women are taller than 1.8 meters. Also, 60% of the students are women. If a student selected at random is found to be taller than 1.8 meters, then the probability that the student being a woman is

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$\frac{6}{11}$

$\frac{5}{11}$

$\frac{8}{11}$

$\frac{3}{11}$

answer is A.

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Detailed Solution

$P (getting a students taller than 1.8 m)$

$P (T) = \frac{40}{100} \times \frac{4}{100} + \frac{60}{100} \times \frac{1}{100} = \frac{16}{1000} + \frac{6}{1000} = \frac{22}{1000} = \frac{11}{500}$

$P (G / T) = \frac{P (T / G) P (G)}{P (T)}$

$= \frac{1}{100} \times \frac{66}{100} / \frac{11}{500}$

$= \frac{6}{1000} / \frac{11}{500} = \frac{3}{11}$

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