So everyone, I have a simple probability question that I cannot wrap my head around. It go like this:
There are $6$ men $9$ women on committee. $2$ people are chosen for treasurer and secretary. What is the probability one man is chosen for treasurer and one woman for secretary?
My solution goes like this:
$6$ men/$15$ people $\cdot$ $9$ women/$14$ people = $9/35$
but I was thinking, if we think that the denominator will be $15$ choose $2 = 105$, then we get different answer of $6 \cdot 9/105 = 18/35$???
Now I'm confused on which one is the correct answer. Someone help clear the confusion for me. Thanks.
Best Answer
Let's call the "first" position treasurer and "second" position secretary. Then our order of choice matters in this setup.
Your initial computation (9/35) is correct since it computes the probability of putting a man in the first position and a woman in the second.
Your next computation (18/35) computes the ratio of the number of ways to choose a man and woman to the total number of ways to choose two people. But the numerator must be halved to account for the fact that only half of those ways place the man in the first position and woman in the second. This results in the correct answer.