# A question in Probability, resignations from stores

"Stores A,B and C have 50,75 and 100 employees and respectively, 30, 60 and 70 percent of these are women. Resignations are equally likely among all employees, regardless of sex. One employee resigns and this is a woman. What is the probability that she works in store C ?"

Source : A First Course in Probability, Sheldon Ross, Chapter 3, Exercise 32

-
add comment

## 1 Answer

Store A has $50\times 0.3=15$ women; Store B has $75\times 0.6=45$ women; and Store C has $100\times 0.7=70$ women.

The probability $P(\mbox{one woman resigns})=\displaystyle\frac{15+45+70}{50+75+100}=\frac{130}{225}.$

The probability $P(\mbox{one woman resigns and the woman is from Store C})=\displaystyle\frac{70}{50+75+100}=\frac{70}{225}.$

The conditional probability $P(\mbox{the woman is from Store C}|\mbox{one woman resigns})$ $$=\displaystyle\frac{P(\mbox{one woman resigns and the woman is from Store C})}{P(\mbox{one woman resigns})}=\frac{\frac{70}{225}}{\frac{130}{225}}=\frac{7}{13}.$$

-
I got the same answer actually. But their answer was 1/2. That's why I asked this question. –  user118102114 Jan 6 '12 at 6:05
@user118102114 If you got this same answer (presumably before posting), then you should have included it with your question. –  Austin Mohr Jan 6 '12 at 6:49
add comment