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I'm having a tough time understanding how combinations and permutations work in complex question.

The question goes like this: If a board of 12 people is to be selected randomly from a pool of 15 people, and the pool consists of 2/3 men and 1/3 women, what is the probability that the board will comprise at least 2/3 men?

I figured out that there are 10 men and 5 women in the pool, and came up with the following equation. 10C8 * 7C4 / 15C12 (choosing 8 from 10 men pool) * ( choosing 3 from remaining 7 people pool) / (choosing 12 from 15 people pool). I argument is that if I choose 8 men and then choose any combination of 4 people, then I will have the board with at least 2/3 men. Sadly, the result of my equation proved that I'm completely misunderstanding this topic :(

Can anybody help me understand what I did wrong, and how I should approach questions like this?

Probably confused
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    You wrote $3$ but meant $4$ from $7$. This unfortunately double-counts some of the committees with more than $8$ men. For example, for comittees with $9$ men, it double-counts the case when $8$ specific men $m_1$ to $m_8$ are in the first bunch, and Charlie is in the second batch, and when Charlie and $m_1$ to $m_7$ are in the first batch and $m_8$ in the second. There is a lot of additional multiple counting. – André Nicolas Aug 19 '15 at 01:39

2 Answers2

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Idea:

It may be easier to think about ways of selecting $3$ people not to be on the board. If the board is to be at least $\frac23$ men, you are interested in cases where the three non-board members include $0$, $1$, or $2$ men.

paw88789
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It's better to look at the people not in the board. Out of those 3 people, you want at least one to be female. The chance of none of them being female is: $$\dfrac{10}{15}*\dfrac{9}{15}*\dfrac{8}{15}=\dfrac{16}{75}$$ And this means that the chance of at least one of the left out people is a woman: $$1-\dfrac{16}{75}=\dfrac{59}{75}$$

Mastrem
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