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 $\frac{19}{72}$
Best Answer
Say you draw from box A first. There are two opportunities: WB and BW, as The Chaz said. BW's chance is
(5/12)*(2/6), or 10/72. WB's chance is(7/12)*(4/6), or 28/72. This is exactly what you got, and when you add then you do get 38/72, or 19/36, which is your answer. Brian is likely right, after a few people trying this and getting the same answer, the only likely possibility is an error in your book.