write your answer as a simplified fraction.

Question

A bag contains 3 red balls, 5 blue balls, and 6 yellow balls.

a) If 3 balls are chosen at random, what is the probability that 2 yellow and 1 red ball

will be selected?

b) If 1 ball is chosen at random, what is the probability that it is red or blue?

c) If 1 ball is chosen at random, what is the probability that it is red, blue, or yellow?

Answer 1

This is again a question of finding the percentage of favorable outcomes against the total universe of outcomes.

a) 3 balls can be chosen out of 14 in 14C3 ways. 2 yellow balls out of 6 can be chosen in 6C2 ways and 1 red ball can be chosen out of 3 in 3C1 ways.

Thus the required probability is (6C2 * 3C1) / (14C3)

b) A ball can be red or blue in 8 (3+5) ways. Thus the required probability would be 8/14

c) Any ball you select will be red, blue or yellow!. Probability is the same as that of a certain event!

Answer 2

I will solve a. b and c are simple.

We have 3 red balls, 5 blue balls, and 6 yellow balls.

P(2 yellow balls and 1 blue ball) = the number of combinations with 2 yellow balls and 1 blue ball)/the number of combinations with 3 balls)

There are 3 + 5 + 6 = 14 balls, 6 of which are yellow and 5 of which are blue.

Thus, the number of combinations with 2 yellow balls of 6 and 1 blue ball of 5/ the number of combinations with 3 balls of 14 =

C(6,2)C(5,1)/C(14,3) =

15*5/364 =

75/364