The probability of men getting a certain dise | vedclass

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Question

(A) Let $M$ be the event that a man is chosen and $W$ be the event that a woman is chosen. Let $D$ be the event that the person has the disease and $D

Vedclass · Accepted Answer

$\frac{75}{107}$

Vedclass · Answer

(A) Let $M$ be the event that a man is chosen and $W$ be the event that a woman is chosen. Let $D$ be the event that the person has the disease and $D^c$ be the event that the person does not have the disease. Let $T^+$ be the event that the blood test is positive.Given:$P(M) = \frac{30}{50} = \frac{3}{5}$$P(W) = \frac{20}{50} = \frac{2}{5}$$P(D|M) = \frac{1}{2}$,so $P(D^c|M) = 1 - \frac{1}{2} = \frac{1}{2}$$P(D|W) = \frac{1}{5}$,so $P(D^c|W) = 1 - \frac{1}{5} = \frac{4}{5}$The test is correct with probability $\frac{4}{5}$,so:$P(T^+|D) = \frac{4}{5}$ (True positive)$P(T^+|D^c) = 1 - \frac{4}{5} = \frac{1}{5}$ (False positive)We want to find $P(M|T^+)$. By Bayes' Theorem:$P(M|T^+) = \frac{P(M) \cdot P(T^+|M)}{P(T^+)}$$P(T^+|M) = P(T^+|D)P(D|M) + P(T^+|D^c)P(D^c|M) = (\frac{4}{5} 	imes \frac{1}{2}) + (\frac{1}{5} 	imes \frac{1}{2}) = \frac{4}{10} + \frac{1}{10} = \frac{5}{10} = \frac{1}{2}$$P(T^+|W) = P(T^+|D)P(D|W) + P(T^+|D^c)P(D^c|W) = (\frac{4}{5} 	imes \frac{1}{5}) + (\frac{1}{5} 	imes \frac{4}{5}) = \frac{4}{25} + \frac{4}{25} = \frac{8}{25}$$P(T^+) = P(M)P(T^+|M) + P(W)P(T^+|W) = (\frac{3}{5} 	imes \frac{1}{2}) + (\frac{2}{5} 	imes \frac{8}{25}) = \frac{3}{10} + \frac{16}{125} = \frac{75 + 32}{250} = \frac{107}{250}$$P(M|T^+) = \frac{\frac{3}{5} 	imes \frac{1}{2}}{\frac{107}{250}} = \frac{3/10}{107/250} = \frac{3}{10} 	imes \frac{250}{107} = \frac{75}{107}$

Box	White	Green	Red	Probability
$A$	$1$	$2$	$3$	$\frac{1}{2}$
$B$	$2$	$3$	$1$	$\frac{1}{3}$
$C$	$3$	$1$	$2$	$\frac{1}{6}$

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