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I am approaching this problem with two ways:

1) Brute force method: physically laying out all possible outcomes and count the outcomes in which there are women than men.

2) Using counting principles: Applying counting principles, combining with laws of probability.

So, my first method goes like this:

  1. I listed out all the possible outcomes ($2^{4}$ outcomes), and I have the following:

{(M, M, M, M), (W, M, M, M), (M, W, M, M), (M, M, W, M), (M, M, M, W), (W, W, M, M), (M, W, W, M), (M, M, W, W), (W, M, M, W), (W, W, W, M), (W, W, M, W), (W, W, W, M), (W, M, W, W), (W, W, M, W), (W, W, W, M), (W, W, W, W)}

So, the probability is obviously 7/16.

  1. Now, I am trying to get the same result using counting principles and probability theorems (and here is I have almost always have difficulty with).

let A be the number of woman in the pool, and B be the number of men in the pool.

So, the probability of picking will be P(A)+ P(B).

Then, I figure that the probability (without the condition of women more than men) will be

$$\frac{1}{5!}+\frac{1}{4!}=\frac{3}{20}$$

Then, I get stuck because it appears that my thinking is not correct. And even if I am half way there, how do I account for the probability of the 16 outcomes for picking the committee members and then focusing on the outcomes that there are more women than men?

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  • $\begingroup$ In your first method, the result is not correct because the outcomes are not equally likely. $\endgroup$ – Daniel Mathias May 16 '20 at 11:01
  • $\begingroup$ I know there is something wrong. Thank you for pointing out. But what is the correct way? $\endgroup$ – Amos Ku May 16 '20 at 16:34
  • $\begingroup$ You need either four women or three women and one man. Use the choose function (binomial distribution) to find the number of qualifying groups, then divide by the number of unrestricted groups. See my comment on Alex's answer. $\endgroup$ – Daniel Mathias May 16 '20 at 16:42
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There are $\binom{9}{4}$ ways to select 4 people out of 9. To get more men than women you need to either get $(4,0), (3,1), (2,2)$. The probability of the compliment is

$$ 1-\bigg(\frac{\binom{5}{4}}{\binom{9}{4}} +\frac{\binom{5}{3}\binom{4}{1}}{\binom{9}{4}} + \frac{\binom{5}{2}\binom{4}{2}}{\binom{9}{4}} \bigg) $$

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    $\begingroup$ Direct count: $\frac{\binom{4}{4}+\binom{4}{3}\binom{5}{1}}{\binom{9}{4}}$ $\endgroup$ – Daniel Mathias May 16 '20 at 11:07

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