Question

A group consists of 15 people: 5 Americans, 6 Canadians and 4 Mexicans.

(a) In how many ways can a sub-group of 4 people be selected?







(b) In how many ways can a sub-group of 4 people consist of all Canadians?







(c) What is the probability that a sub-group will consist of all Canadians?







(d) In how many ways can a sub-group of 4 people consist of 2 Canadians and 2 Mexicans?







(e) In how many ways can a sub-group of 4 consist of 1 American, 1 Mexican and 2 Canadians?

Answer 1

A group consists of 15 people: 5 Americans, 6 Canadians and 4 Mexicans.

(a) In how many ways can a sub-group of 4 people be selected?

(b) In how many ways can a sub-group of 4 people consist of all Canadians?

(c) What is the probability that a sub-group will consist of all Canadians?

(d) In how many ways can a sub-group of 4 people consist of 2 Canadians and 2 Mexicans?

(e) In how many ways can a sub-group of 4 consist of 1 American, 1 Mexican and 2 Canadians?

Solution

(a) In how many ways can a sub-group of 4 people be selected?

(5+6+4 people choose 4)=(15 people choose 4)=C(15,4)=1365 ways

(b) In how many ways can a sub-group of 4 people consist of all Canadians?

(6 Canadians choose 4)=C(6,4)=15 ways

(c) What is the probability that a sub-group will consist of all Canadians?

P(all Canadians) = $\frac{6}{15} * \frac{5}{14} * \frac{4}{13} * \frac{3}{12} * \frac{2}{11} * \frac{1}{10} = 0.0002$

(d) In how many ways can a sub-group of 4 people consist of 2 Canadians and 2 Mexicans?

(6 Canadians choose 2) times (4 Mexicans choose 2)=C(6,2)*C(4,2)=15*6=90 ways

(e) In how many ways can a sub-group of 4 consist of 1 American, 1 Mexican and 2 Canadians?

(5 Americans choose 1) times (4 Mexicans choose 1) times (6 Canadians

choose 2)=C(5,1)*C(4,1)*C(6,2)=5*4*15=300 ways