Probability in urns There are 3 urns labeled X, Y, and Z.


*

*Urn X contains 4 red balls and 3 black balls.

*Urn Y contains 5 red balls and 4 black balls.  

*Urn Z contains 4 red balls and 4 black balls.


One ball is drawn from each of the 3 urns. What is the probability that, of the 3 balls drawn, 2 are red and1 is black?
I could not understand this problem properly and find out how solution is obtained.
 A: Add up the probabilities of all mutually exclusive cases


*

*RRB  $\ \longrightarrow \ \dfrac{4}{7}\cdot \dfrac{5}{9}\cdot \dfrac{4}{8}$

*RBR etc

*BRR etc

A: You want to find
$$\mathsf P\Big((X{=}r\cap Y{=}r\cap Z{=}b)\cup(X{=}r\cap Y{=}b\cap Z{=}r)\cup(X{=}b\cap Y{=}r\cap Z{=}r)\Big)$$
Which is the probability of the union of disjoint events each formed from the intersection of independent events.
A: If the three expressions $r+r+r+r + b+b+b$, $r+r+r+ r+r+b+b+b+b$, and $b+b+ b+b + r+r+r+r$ are multiplied to obtain a polynomial in $r$ and $b$, the coefficient of $b^i r^j$ represents the number of ways to choose one ball from each urn and obtain $i$ black and $j$ red balls. So the number of ways to obtain two red and one black is the coefficient of $r^2b$ in $(4r+3b) (5r+4b) (4r+4b)=48 b^3 + 172 b^2 r + {\color{red}{204 b r^2}} + 80 r^3$, or 204. The total number of possible outcomes with any result is the $7\times8\times9=504$, or equivalently the sum of the coefficients of the expanded polynomial. So the probability is $\dfrac{204}{504}$.
