Question

Rework problem 7 from section 3.4 of your text, involving the selection of a colored ball from one of three bowls. Assume that you randomly pick one of the bowls: X, Y, or Z. You then randomly draw one ball out of the selected bowl and note its color. The bowls contain the following colored balls: Bowl X: 3 red, 2 white, and 1 blue. Bowl Y: 2 red, 1 white, and 2 blue. Bowl Z: 2 red, 1 white, and 2 blue.

(1) If the ball is blue, what is the probability that it was drawn from bowl X?

(2) If the ball is either blue or white, what is the probability that it was drawn from bowl Y?

Answer 1

1)P(blue) =P(X and blue ball)+P(Y and blue ball)+P(Z and blue ball)

=(1/3)*(1/6)+(1/3)*(2/5)+(1/3)*(2/5)=29/90

therefore P(X given blue)=P(X and blue ball)/P(blue)=(1/3)*(1/6)/(29/90)=5/29

2)

P(either blue or white)=P(X and either blue or white)+P(Y and either blue or white)+P(Z and either blue or white)

=(!/3)*(3/6)+(1/3)*(3/5)+(1/3)*(3/5)=17/30

hence P(Y given either blue or white)=P(Y and either blue or white)/P(either blue or white)

=(1/3)*(3/5)/(17/30)=6/17