Question

How would i solve a problem like this?

i'm confused on how probability works with combinations, could someone explain using this example? thanks!

Answer 1

The number of combinations of size
`k` that you can make with
`n` items is given by the so-called binomial coefficient,

`\dbinom nk = (n!)/(k!(n-k)!)`

•
`n!` is the number of ways of permuting
`n` items.

•
`(n-k)!` is the number of ways of permuting all but
`k` of the
`n` items.

Dividing
`n!` by
`(n-k)!` then gives the number of ways of permuting only
`k` of the total
`n` items.

•
`k!` is the number of ways of permuting
`k` items.

Dividing
`(n!)/((n-k)!)` by
`k!` then removes all those permutations which contain the same items. We call these combinations.

For this problem we only care about counting combinations.

There are

`\dbinom 63 = (6!)/(3!(6-3)!) = 20`

ways of selecting any 3 girls from the total 6 girls in the entire group of people.

There are

`\dbinom 72 = (7!)/(2!(7-2)!) = 21`

was of selecting any 2 boys from the total 7 boys.

Then there are

`\dbinom 63 \dbinom 72 = 20\cdot21 = \boxed{420}`

ways of choosing a committee of 5 people consisting of 3 girls and 2 boys.

If the next question were, "What is the probability that a committee of 5 randomly selected people consists of 3 girls and 2 boys?", then you would additionally need to compute the number of ways one can make a committee of 5 people from the total 13, which is

`\dbinom{13}5 = (13!)/(5!(13-5)!) = 1287`

Then the probability of selecting such a committee at random is

`\frac{\binom 63 \binom72}{\binom{13}5} = (420)/(1287) = (140)/(429) \approx 0.3263`