Question

You have a bag with 21 balls: 7 black, 7 yellow, 7 green

You are picking 5 WITHOUT replacement.

a) The probability of picking four of the same color and one different

b) The probability of picking all that are the same color

c) The probability of having one of each color from the five picked.

Answer 1

a) P(Four of same color and 1 different)

= P(4 black, 1 other) + P(4 yellow, 1 other) + P(4 Green, 1 other)

= (7/21)*(6/20)*(5/19)*(4/18)*(1/14) + (7/21)*(6/20)*(5/19)*(4/18)*(1/14) + (7/21)*(6/20)*(5/19)*(4/18)*(1/14)

= 1/798

= 0.00125

b) P(All of same color)

= P(All black) + P(All Yellow) + P(All Green)

= (7/21)*(6/20)*(5/19)*(4/18)*(3/17) + (7/21)*(6/20)*(5/19)*(4/18)*(3/17) + (7/21)*(6/20)*(5/19)*(4/18)*(3/17)

= 1/323

= 0.00310

c) For this problem, we can apply reverse approach.

P(Having one of each color)

= 1 - [P(Having all balls of same color) + P(Balls consisting of 2 colors)]

= 0.70175