Question

In a group of people, 32 are Australian-born, 12 were born in Asia and 7 were born in

Europe. If 2 of the people are selected at random, find the probability that:
a they were both born in Asia
b at least 1 of them will be Australian-born
с both were born in Europe.

Answer 1

a. 22/425

b. 368/425.

c. 7/425.

Explanation:

There are a total of 32+12+7 = 51 people in the group.

a. Probability (First one is Asian) = 12/51 = 4/17.

Probability(second one is Asian) = 11/50.

So the probability that both are Asian is 4/17 * 11/50

= 22/425.

b. Prob(First is Australian) = 32/51

Prob( Second is not Australian) = 19/50.

Prob( first is Australian and Second is not) = 32/51 * 19/50.

Prob( First is not Australian and the second is) = 19/51 * 32/50.

Prob ( Both are Australian) = 32/51 * 31/50)

Prob( At least one is Australian = 32/51 * 19/50 + 19/51 * 32/50 + 32/51 * 31/50

= 368/425.

c. Prob ( First choice born in Europe) = 7/51 and probability second was also born in Europe = 6/50

So the answer is 7/51 * 6/50 = 7/425.