Question 1134972: 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: 2 red, 1 white, and 2 blue.
Bowl Y: 1 red, 3 white, and 4 blue.
Bowl Z: 2 red, 3 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 by greenestamps(13215)   (Show Source):
You can put this solution on YOUR website!
(1) If the ball is blue, what is the probability that it was drawn from bowl X? That probability is
P((blue) AND (drawn from X))
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P(blue)
P((blue) AND (drawn from X)) = (1/3)(3/6) = 1/6 A (equal probability of selecting any of the 3 bowls; 3 of the 6 balls in X are blue)
P((blue) AND (drawn from Y)) = (1/3)(4/8) = 1/6 B (equal probability of selecting any of the 3 bowls; 4 of the 8 balls in Y are blue)
P((blue) AND (drawn from Z)) = (1/3)(2/7) = 2/21 C (equal probability of selecting any of the 3 bowls; 2 of the 7 balls in Z are blue)
P(blue) = 1/6 + 1/6 + 2/21 = (7+7+4)/42 = 18/42 = 3/7 D (sum of A, B, and C)
P(blue AND drawn from X, given blue) = (1/6)/(3/7) = (1/6)*(7/3) = 7/18 (A/D)
(2) If the ball is either blue or white, what is the probability that it was drawn from bowl Y?
Exactly the same kind of problem; you can solve it using the same method.
P((blue OR white) AND (drawn from X)) = A
P((blue OR white) AND (drawn from Y)) = B
P((blue OR white) AND (drawn from Z)) = C
P(blue OR white) = D = A+B+C
P(blue OR white AND drawn from Y, given blue or white) = B/D
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