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A class consist of 7 females and 12 males.if a committee of 5 is chosen at random for the class,obtain the probability that:
1)5 males are selected
2)exactly 4 males are selected
3)at least 1 female is selected
4)exactly 3 females are selected.

Ezeaku21:
A class consist of 7 females and 12 males.if a committee of 5 is chosen at random for the class,obtain the probability that:
1)5 males are selected
2)exactly 4 males are selected
3)at least 1 female is selected
4)exactly 3 females are selected.

This is probability involving permutations and combinations
For solution:
There are a total of 19 people in the class (7+12). So the general probability of selecting a committee of 5 from 19 is 19C5 = 19!/(19-5)!*5! = 19!/(14!*5!). I'm sure you know ! means factorial, e.g. n! = n * (n-1)*(n-2)*(n-3)*..*1
You should be able to work those values out for 19!, 14! and 5!. The result above will always be the denominator for the probability calculations in all the questions from (1) to (4). Having said that,
(1) Prob of 5 males selected means 5 males selected from 12 males and no female selected, as it's a committee of 5 .
Prob = 12C5/19C5....Work this out
(2) Prob of exactly 4 males selected means 4 males selected from 12 and 1 female selected from 7
= (12C4 * 7C1)/19C5
(3) Prob of at least 1 female means 1 female and above, i.e. 1 F and 4 M or 2 F and 3M or 3F and 2M or 4F and 1 M or 5F and 0M ( where F = female and M = male)
Prob = (7C1*12C4)/19C5 + (7C2 * 12C3)/19C5 + (7C3 * 12C2)/19C5 + (7C4 * 12C1)/19C5 + 7C5/19C5
(4) Prob of exactly 3 females means 3 females and 2 males
= (7C3 * 12C2)/19C5
That's all.. Goodluck