Question

A car dealership is giving away a trip to Rome to one of their 120 best customers. In this group, 65 are women, 80 are married and 45 married women. If the winner is married, what is the probability that it is a woman?

Answer 1

Let $W$ denote the event " winner is a women". Then $W'$ denote the event "winner is not a men".

Let $M$ denote the event "winner is married". Then $M'$ denote the event "winner is not married".

$$P(W)=\dfrac{65}{120}=\dfrac{13}{24}$$

$$P(W')=1-P(W)=1-\dfrac{13}{24}=\dfrac{11}{24}$$

$$P(W)=\dfrac{65}{120}=\dfrac{13}{24}$$

$$P(W')=1-P(W)=1-\dfrac{13}{24}=\dfrac{11}{24}$$

$$P(M|W)=\dfrac{45}{65}=\dfrac{9}{13}$$

$$P(M|W')=\dfrac{80-45}{120-65}=\dfrac{7}{11}$$

By Bayes' Theorem

$$P(W|M)=\dfrac{P(M|W)P(W)}{P(M|W)P(W)+P(M|W')P(W')}$$

$$P(W|M)=\dfrac{\dfrac{9}{13}(\dfrac{13}{24})}{\dfrac{9}{13}(\dfrac{13}{24})+\dfrac{7}{11}(\dfrac{11}{24})}=\dfrac{9}{16}$$

If the winner is married, the probability that it is a woman is $\dfrac{9}{16}.$