Question

For this problem, assume that three students are chosen from 5 freshmen, 4 sophomores, and 4 juniors.

What is the probability of selecting 2 freshmen and 1 junior given that at least one freshman will be selected?

Answer 1

The formula $C(n, r)= \frac{n!}{r!(n-r)!}$, where r! is 1*2*3*...r

is the formula which gives us the total number of ways of forming groups of r objects, out of n objects.

for example, given 10 objects, there are C(10,6) ways of forming groups of 6, out of the 10 objects.

The sample space S in our problem is as follows: 

triples selected from 5 freshmen, 4 sophomores, and 4 juniors, so that at least 1 freshman is selected.

The selection of 1 freshman can be done in 5 ways. The selection of the remaining 2 out of 4 freshmen (one is already chosen), 4 sophomores, and 4 juniors can be done in C(12, 2) many ways.

$C(12, 2)=\displaystyle{ \frac{12!}{2!10!}= \frac{12\cdot11\cdot10!}{2!10!}=66$

So there are in total, 5*66= 330 possible triples.

n(S)=330

The event E "selecting 2 freshmen and 1 junior" can occur in :

$C(5, 2)\cdot 4=\displaystyle{ \frac{5!}{2!3!}\cdot4=40$ many ways.

so n(E)=40

P(E)=n(E)/n(S)=40/330=0.12

Answer: 0.12