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38 votes
How would i solve a problem like this?

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

How would i solve a problem like this? i'm confused on how probability works with-example-1
User Willian Fuks
by
2.9k points

1 Answer

20 votes
20 votes

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

User PeterSW
by
2.5k points
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