Question

In a class of 19 students, 3 are math majors. A group of four students is chosen at random. (Round your answers to four decimal places.) (a) What is the probability that the group has no math majors? (b) What is the probability that the group has at least one math major? (c) What is the probability that the group has exactly two math majors?

Answer 1

Answer:

(a) The probability is 0.4696

(b) The probability is 0.5304

(c) The probability is 0.0929

Step-by-step explanation:

The total number of ways in which we can select k elements from a group n elements is calculate as:

$nCx=\frac{n!}{x!(n-x)!}$

So, the number of ways in which we can select four students from a group of 19 students is:

$19C4=\frac{19!}{4!(19-4)!}=3,876$

On the other hand, the number of ways in which we can select four students with no math majors is:

$(16C4)*(3C0)=(\frac{16!}{4!(16-4)!})*(\frac{3!}{0!(3-0)!})=1820$

Because, we are going to select 4 students form the 16 students that aren't math majors and select 0 students from the 3 students that are majors.

At the same way, the number of ways in which we can select four students with one, two and three math majors are 1680, 360 and 16 respectively, and they are calculated as:

$(16C3)*(3C1)=(\frac{16!}{3!(16-3)!})*(\frac{3!}{1!(3-1)!})=1680$

$(16C2)*(3C2)=(\frac{16!}{2!(16-2)!})*(\frac{3!}{2!(3-1)!})=360$

$(16C1)*(3C3)=(\frac{16!}{1!(16-1)!})*(\frac{3!}{3!(3-3)!})=16$

Then, the probability that the group has no math majors is:

$P=\frac{1820}{3876} =0.4696$

The probability that the group has at least one math major is:

$P=\frac{1680+360+16}{3876} =0.5304$

The probability that the group has exactly two math majors is:

$P=\frac{360}{3876} =0.0929$