A committee of three people is selected at random from a collection of people, seven from Canada, five from Mexico, and three from the United States.

Question

(a) What is the probability that the committee consists entirely of Canadians?

(b) What is the probability that the committee has no Canadians at all?

Answer 1

Hello!

To solve this we need the total number of people, which totals to 15. Of this we have 7 Canadians, 5 Mexicans, and 3 Americans.

(a)

To find the probability of selecting the first committee member as a Canadian we take the number of Canadians divided by the total number:

7 / 15 = 0.467 or 46.7%

The second Canadian randomly selected has a different probability because we have to remove the first member selected from the numbers. So now there are only 6 Canadians of 14 people.

6 / 14 = 0.428 or 42.8%

The third is similar:

5 / 13 = 0.385 or 38.5%

Now to find the probability of doing ALL of these things, we multiply them together.

Total: 0.467 * 0.428 * 0.385 = 0.077 or 7.7%

(b)

This problem follows the same solution, but with the OTHER members so our first probability is 8/15.

8 / 15 = 0.533

7 / 14 = 0.5

6 / 13 = 0.462

Total: 0.533 * 0.5 * 0.462 = 0.123 or 12.3%