Conditional probability: take a ball from the first box

There are 3 white balls and 2 black balls in the first box. There are 4 white and 4 black balls in the second box. A ball is randomly chosen from the first box and placed in the second box. A ball is then randomly chosen from the second box. What's the probability the chosen ball is white?

Here's my attempt:

• A - a white ball was chosen from the first box and placed in the second box
• B - a white ball was chosen from the second box

$P(A)=3/5$ because 5 balls of which 3 are white

$P(B|A)=5/9$ because in the 2nd box there are now 9 balls (4 white, 4 black + the chosen one from the 1st box which was white)

$P(B|\bar{A})=4/9$ because in the 2nd box the ninth ball is black

I'm calculating the probability of event B using this formula:

$P(B)=P(B|A)*P(A)+P(B|\bar{A})*P(\bar{A})=5/9*3/5+4/9*2/5=23/45$

Is my solution correct?

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looks okay to me –  Aang Mar 7 '13 at 10:11