Help finding probability

**benny92000** · August 14th 2011, 03:19 PM

Here is the problem: A committee of 5 people is to be selected from 6 men an women. If the selection is made randomly, what is the probability that the committee consists of 3 men and 2 women?

The answer is 1260/3003. Help is appreciated. I know the formulas for combinations/permutations, but I don't know how to apply them here.

**pickslides** · August 14th 2011, 03:59 PM

From your question I would try (6C3+6C2)/12C5, but this won't give the answer you need.

Have you left any details out?

**aonin** · August 14th 2011, 04:08 PM

Originally Posted by benny92000

Here is the problem: A committee of 5 people is to be selected from 6 men an women. If the selection is made randomly, what is the probability that the committee consists of 3 men and 2 women?

The answer is 1260/3003. Help is appreciated. I know the formulas for combinations/permutations, but I don't know how to apply them here.

If the number of men and women before they are selected are known then the probability is $P=\frac{\text{(number of men C 3)}\times \text{(number of women C 2)}}{6 C 5}$

this is a combinations question, as the order that people are selected do not matter, but this doesn't get the answer you need.

**downthesun01** · August 14th 2011, 08:59 PM

I'm confused are there 12 people total - 6 males and 6 females? Or are there 6 total people, and it's unknown how many are male or female?

If it's 12 total people then the answer should be

$\frac{{{6}\choose{3}} {{6}\choose{2}}}{{{12}\choose{5}}}$

But that doesn't give the answer you stated.

If the case is that there are 6 people and their genders are unkown, then

The answer should be

$\frac{{{x}\choose{3}} {{6-x}\choose{2}}}{{{6}\choose{5}}}$

Where x is the number of males in the group.

x can only be 3 or 4, and neither gives the answer you stated.

Are you should you're stating the question correctly?

**benny92000** · August 15th 2011, 06:47 AM

I'm sorry guys. 6 men and 9 women... downthesun, what do those parentheses notations man? my book uses them but it never explained them.

**Plato** · August 15th 2011, 06:58 AM

Originally Posted by benny92000

I'm sorry guys. 6 men and 9 women... downthesun, what do those parentheses notations man? my book uses them but it never explained them.

Fifteen, choose five.
$_{15}\mathcal{C}_5=\binom{15}{5}=\frac{15!}{5!(10! )}$

**aonin** · August 16th 2011, 04:56 AM

Originally Posted by benny92000

I'm sorry guys. 6 men and 9 women... downthesun, what do those parentheses notations man? my book uses them but it never explained them.

Since its 6 men and 9 women the probability is
$P=\frac{\text{no. of ways of choosing 3 men from 6 men}\times\text{no. of ways of choosing 2 women from 9 women}}{\text{total no. of ways of choosing 5 people from 15 people}}\linebreak P=\frac{6C3\times9C2}{15C5}=\frac{720}{3003}$

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