A meeting has 12 employees. Given that 8 of the employees is a woman, find the probability that all the employees are women?(a) \(\frac{11}{23}\)(b) \(\frac{12}{35}\)(c) \(\frac{2}{9}\)(d) \(\frac{1}{8}\)I had been asked this question in a national level competition.This interesting question is from Discrete Probability topic in division Discrete Probability of Discrete Mathematics

A meeting has 12 employees. Given that 8 of the employees is a woman,

1.	A meeting has 12 employees. Given that 8 of the employees is a woman, find the probability that all the employees are women?(a) \(\frac{11}{23}\)(b) \(\frac{12}{35}\)(c) \(\frac{2}{9}\)(d) \(\frac{1}{8}\)I had been asked this question in a national level competition.This interesting question is from Discrete Probability topic in division Discrete Probability of Discrete Mathematics
Answer» The correct ANSWER is (c) \(\frac{2}{9}\) To explain I would say: Assume that the probability of an EMPLOYEE being a man or WOMAN is (\(\frac{1}{2}\)). By using Bayes’ theorem: LET B be the event that the meeting has 3 employees who is a woman and let A be the event that all employees are WOMEN. We want to find P(A\|B) = \(\frac{P(B\|A)P(A)}{P(B)}\). P(B\|A) = 1, P(A) = \(\frac{1}{12}\) and P(B) = \(\frac{8}{12}\). So, P(A\|B) = \(\frac{1\frac{1}{12}}{\frac{8}{12}} = \frac{1}{8}\).

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