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In box A there are $7$ white balls and $5$ black balls. In box B there are $2$ white balls and $4$ black balls. One ball is drawn from each box. What is the probability of getting one ball of each color?

In order to find the possible cases I multiplied the 12 balls in the box A by the $6$ balls in box B. There are $72$ possible cases. Then I set $2$ different events:

A-"draw a ball from box A".

W-"draw a white ball".

Then I thought, there are $2$ situations where the drawn balls can have different colors.I can remove a black ball from box A and a white ball from box B or the inverse.

$P(\bar{W}|A) \cdot P(W|\bar{A})=\frac{5}{12}\cdot \frac{2}{6}=\frac{10}{72}$

$P(W|A) \cdot P(\bar{W}|\bar{A})=\frac{7}{12} \cdot \frac{4}{6}=\frac{28}{72}$

Then I added the two probabilities:$\frac{38}{72}$

But the result don't mix with the book solutions. Who is wrong? Thanks. The book solution is $\frac{19}{72}$

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Your answer is fine. My best guess is that whoever wrote up the book’s answer meant to reduce $\frac{38}{72}$ to $\frac{19}{36}$ and forgot to change the denominator. –  Brian M. Scott Jan 30 '12 at 22:41

1 Answer 1

up vote 0 down vote accepted

$72$ cases ?!?! I would go about it in a different way:

There are two cases for box A: black and white.

There are two cases for box B: black and white.

There are four cases total: BB, BW, WB, WW... let's say that the first letter is the color from box A and the second is the color from box B.

Case BB has probability $$\dfrac{5}{12} \cdot \dfrac{4}{6},$$ the product of the probabilities of getting black in A and B.

"One ball of each color" can happen in case BW and in case WB.

Calculate those probabilities and add them.

As for the book's answer, Brian's explanation in the comments is my best guess as to what happened.

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If you set up a double entry table, you can find the favorable cases, by crossing each column with each row.If you sum all favorable cases you end up with the total possible cases. –  João Jan 30 '12 at 23:34

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