Conditional probability problem

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Homework Statement

Urn I contains 3 white and 5 red balls, whereas urn II contains 2 white and 1 red ball. A ball is randomly chosen from urn I and put into urn II, and a ball is then randomly selected from urn II.

(a) What is the probability that the ball selected from urn II is white?

(b) What is the conditional probability that the transferred ball was white, given that a white ball is selected from urn II?

The Attempt at a Solution

for part a) there are two cases. case 1 is that the ball taken from urn 1 was white and case 2 is that the ball taken from urn 1 was red.

case 1:
the first step is to pick the first ball (from urn 1). in this case it was white which has a 3/8 probability of happening
the next step is to pick the ball from urn 2. since the transferred ball was white, the probability of this is now 3/4 so in this case the probability that the first ball was white and the second ball was white is (3/8)(3/4) = 9/32

case 2:
the first step is picking a red ball from urn 1 which has a probability of 5/8. next step is to pick a white ball from urn 2 which since the transferred ball was red there is now a 2/4 probability of. So the probability that the first ball was red and the second ball was white is (5/8)(2/4) = 10/32

so the total probability that the ball from urn 2 is white is the sum of both case:
9/32 + 10/32 = 19/32

b) the probability that the ball taken from urn 1 was white given that the ball taken from urn 2 is white is
equal to (probability that the ball from urn 1 was white and the ball from urn 2 was white)/(probability that the ball from urn 2 was white) = (9/32)/(19/32) = 9/19

is my reasoning correct?

 

[RUber]

Yep.

 

[toothpaste666]

Thank you!