Question

in a group of 13 students, there are 6 females and 7 males. how many different 3 - member committees are possible if there are no restrictions? how many different 3 - member committees are possible there must be exactly 2 girls? how many different 3 - member committees are possible there must be at least 1 girl? determine the probability that a 3 - 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 $n = 13$ (total students) and $r=3$ (committee - size).
$C(13,3)=\frac{13!}{3!(13 - 3)!}=\frac{13!}{3!×10!}=\frac{13\times12\times11}{3\times2\times1}=286$

Step2: Calculate committees with exactly 2 girls

First, choose 2 girls out of 6: $C(6,2)=\frac{6!}{2!(6 - 2)!}=\frac{6\times5}{2\times1}=15$. Then choose 1 boy out of 7: $C(7,1)=\frac{7!}{1!(7 - 1)!}=7$. By the multiplication principle, the number of 3 - member committees with exactly 2 girls is $C(6,2)\times C(7,1)=15\times7 = 105$

Step3: Calculate committees with at least 1 girl

Find the number of committees with no girls (all boys) first. $C(7,3)=\frac{7!}{3!(7 - 3)!}=\frac{7\times6\times5}{3\times2\times1}=35$. Then subtract from the total number of committees with no restrictions. The number of committees with at least 1 girl is $C(13,3)-C(7,3)=286 - 35=251$

Step4: Calculate the probability of at least 1 girl

The probability $P$ of having at least 1 girl in a 3 - member committee is $\frac{\text{Number of committees with at least 1 girl}}{\text{Total number of committees}}$. So $P=\frac{251}{286}\approx0.878$. As a percentage, $P = 87.8\%$

Answer:

286
105
251
87.8%