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The problem I am working on is:

One box contains six red balls and four green balls, and a second box contains seven red balls and three green balls. A ball is randomly chosen from the first box and placed in the second box. Then a ball is randomly selected from the second box and placed in the first box.

a.What is the probability that a red ball is selected from the first box and a red ball is selected from the second box?

b.At the conclusion of the selection process, what is the probability that the numbers of red and green balls in the first box are identical to the numbers at the beginning?

My attempt at part a:

A = choose red ball from box # 1 $P(A)= 6/10=.6$

B = choose red ball from box # 2

So, if it is the case that a red ball was chosen from the first box, then that would increase the total amount of contents of the 2nd box to 11; consequently, it would then be more likely to choose a red ball from the 2nd box.

$P(B|A) = \frac{8/11}{.6} \approx 1.21$

Obviously, this is incorrect, what have I done wrong?

share|improve this question
You don’t want $P(B\mid A)$: you want $P(A\text{ and }B)$, so you should be multiplying: $$\frac6{10}\cdot\frac8{11}=\frac{24}{55}\;.$$ –  Brian M. Scott Feb 4 '13 at 20:55
@BrianM.Scott How did you generate that grey back-ground? –  Mack Feb 4 '13 at 20:55
With a leading >. Click on edit, and you can see the current underlying code. –  Brian M. Scott Feb 4 '13 at 20:57
@BrianM.Scott Why exactly is it just a simple multiplication? –  Mack Feb 4 '13 at 21:02
Once A has occurred, there are $8$ red and $3$ green balls in the second box, so it’s simply $8/11$. (I’ve a feeling that you’re trying to make it harder than it actually is!) –  Brian M. Scott Feb 4 '13 at 21:12
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