Question

in a group of 12 students, there are 7 females and 5 males. how many different 4 - member committees are possible if there are no restrictions? how many different 4 - member committees are possible there must be exactly 2 girls? how many different 4 - member committees are possible there must be at least 1 girl? determine the probability that a 4 - member committee chosen at random from this group will consist of at least 1 females. write your answer as a percent.

Explanation:

Step1: Calculate committees with no restrictions

Use combination formula $C(n,r)=\frac{n!}{r!(n - r)!}$, where $n = 12$ (total students) and $r=4$ (committee - size).
$C(12,4)=\frac{12!}{4!(12 - 4)!}=\frac{12\times11\times10\times9}{4\times3\times2\times1}=495$

Step2: Calculate committees with exactly 2 girls

Choose 2 girls out of 7 and 2 boys out of 5. Use the multiplication principle of counting.
Number of ways to choose 2 girls: $C(7,2)=\frac{7!}{2!(7 - 2)!}=\frac{7\times6}{2\times1}=21$
Number of ways to choose 2 boys: $C(5,2)=\frac{5!}{2!(5 - 2)!}=\frac{5\times4}{2\times1}=10$
Total number of 4 - member committees with exactly 2 girls is $C(7,2)\times C(5,2)=21\times10 = 210$

Step3: Calculate committees with at least 1 girl

Find the number of committees with no girls (all - boys) and subtract from the total number of committees with no restrictions.
Number of all - boys committees: $C(5,4)=\frac{5!}{4!(5 - 4)!}=5$
Number of committees with at least 1 girl: $C(12,4)-C(5,4)=495 - 5=490$

Step4: Calculate probability of at least 1 female

Probability $P=\frac{\text{Number of committees with at least 1 female}}{\text{Total number of committees with no restrictions}}$
$P=\frac{490}{495}=\frac{98}{99}\approx0.9899$
As a percent, $P\approx98.99\%$

Answer:

No restrictions: 495
Exactly 2 girls: 210
At least 1 girl: 490
Probability of at least 1 female: 98.99%