step by step solution

Question

An urn contains 12 balls: 5 red, 4 white, and 3 blue. A trial consists of selecting one ball from the urn and replacing the ball. Compute the probability that in 3 trials all three balls are different in color?

Answer 1

Count the ways it could happen:

P(RWB) 5/12 x 4/12 x 3/12 +

P(RBW) = 5/12 x 3/12 x 4/12 +

P(BRW) =4/12 x 5/12 x 3/12

P(BWR) = 4/12 x 3/12 x 5/12 +

P(WBR) = 3/12 x 4/12 x 5/12

P(WRB) = 3/12 x 5/12 x 4/12

That's 6 x 3/12 x 4/12 x/5/12

=3/2 x 4/12 x 5/12 = 3/2 x 20/144 = 30/144 = 15/72

15 out of 72 times you'll get 3 different colors