Question

There are 5 blue marbles, 7 red marbles, and 6 white marbles in a bag. Five of the marbles are selected without replacement. What is the probability of selecting two marbles from one color, two marbles from another color, and one from the remaining color?

Answer 1

Number of ways to select r items from n, nCr = n!/(r! x (n-r)!)

Total number of marbles = 5 + 7 + 6 = 18

Number of ways to select 5 marbles = 18C5 = 8,568

P(two marbles from one color, two marbles from another color, and one from the remaining color) = P(2 blue, 2 red and 1 white) + P(2 blue, 1 red and 2 white) + P(1 blue, 2 red and 2 white)

= (5C2 x 7C2 x 6C1 / 18C5) + (5C2 x 7C1 x 6C2 / 18C5) + (5C1 x 7C2 x 6C2 / 18C5)

= (10 x 21 x 6 / 8,568) + (10 x 7 x 15 / 8,568) + (5 x 21 x 15 / 8,568)

= 3,885/8,568

= 0.4534