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 $\frac{19}{72}$