Urn A contains $x$ red marbles and $y$ white marbles, and Urn B contains $z$ red marbles and $v$ white marbles.
If a marble is drawn from Urn A and put into Urn B and then a marble is drawn from Urn B, what is the probability that the second marble is red?
I get the answer $\frac{2z+1}{z+v+1}$, but my book gives the answer $\frac{xz+x+yz}{(x+y)(z+v+1)}$. I cannot understand how the books answer was obtained.
Best Answer
There are two cases: A red marble has been drawn from urn A and a white marble has been drawn from urn A.
$\texttt{First case}$:
The probability that a red marble has been drawn from urn A is $\frac{x}{x+y}$. Then you have $z+1$ red marbles in urn B and $z+v+1$ marbles in total. The probability to draw a red marble from urn B then is $\frac{z+1}{z+v+1}$
$\texttt{Second case}$:
The probability that a white marble has been drawn from urn A is $\frac{y}{x+y}$. Then you have still $z$ red marbles in urn B and $z+v+1$ marbles in total. The probability to draw a red marble from urn B then is $\frac{z}{z+v+1}$
Therefore the probability that hat the second marble is red is
$$\frac{x}{x+y}\cdot \frac{z+1}{z+v+1}+\frac{y}{x+y}\cdot \frac{z}{z+v+1}=\frac{x(z+1)+yz}{(x+y)\cdot (z+v+1)}=\frac{xz+x+yz}{(x+y)\cdot (z+v+1)}$$
Remark on the comment.
Firstly we define the relevant events:
$A \texttt{ :Marble from urn A is red}$
$\overline A\texttt{ :Marble from urn A is white}$
$B \texttt{ :Marble from urn B is red}$
$\overline B\texttt{ :Marble from urn B is white}$
There are two ways where the seocond marble is red.
$\texttt{Marble from urn A is red} \rightarrow \texttt{Marble from urn B is red}$
$\texttt{Marble from urn A is white} \rightarrow \texttt{Marble from urn B is red}$
Compared to only one way you have a higher chance to get a red marble from urn B. Thats why you add the two cases.
If you go along a path each node represents an event, which happen with a specific probability. At path 1 the event A and event B have to happen at the same time. That´s why we multiply the probabilities: $P(A)\cdot P(B|A)$. The second probability is a conditional probability since the probability depends on the what we have drawn from the first urn.
Similar for the second path: $P(\overline A)\cdot P(B|\overline A)$
It is an application of the law of total probability $P(B)=P(A)\cdot P(B|A)+P(\overline A)\cdot P(B|\overline A)$