Urn 1 contains 5 white and 6 black balls, while urn
2 contains 8 white and 10 black balls. Two balls
are randomly selected from urn 1 and are put into
urn 2. If 3 balls are then randomly selected from
urn 2, compute the expected number of white balls
in the trio.
Then I used a hint given by the book:
Let $X_i = 1$ if the i-th white ball initially in urn
1 is one of the three selected, and let $X_i = 0$ otherwise.
Similarly, let $Y_i = 1$ if the i-th white ball from
urn 2 is one of the three selected, and let $Yi = 0$
otherwise. The number of white balls in the trio
can now be written as
$$\sum_{k=1}^5 X_i+\sum_{k=1}^8 Y_i$$
The expected number I'm looking for is
$$\sum_{k=1}^5 E[X_i]+\sum_{k=1}^8 E[Y_i]$$
To calculate $E[X_i]$ if I call $B_1$ the event that the i-th white ball initially in urn 1 is taken and put in the urn 2 and then $B_2$ the event that this white ball initially in urn 1 is one of the three selected
$E[X_i]=p(B_2 B_1)$ I think $B_1$ and $B_2$ are dependent so I can't divide in the product of $p(B_1)*p(B_2)$
I think also that $E[Y_i]=8/20$
Could someone help me?
Best Answer
Just to suggest another solution (based on linearity of expectations), as the book hints, let's compute $P\{Y_i = 1\}$: it's just the probability of selecting the given ball with any two others, which is $\frac{\binom{1}{1}\binom{19}{2}}{\binom{20}{3}} = \frac{3}{20}$.
Similarly, for $B$ being "the $i$th white ball in urn 1", $$ P\{X_i = 1\} = P\{B \text{ is chosen from urn 1} \land B \text{ is chosen from urn 2 after transfer} \} = P\{B \text{ is chosen from urn 2} \mid B \text{ is chosen from urn 1}\} P\{B \text{ is chosen from urn 1}\} = \frac{3}{20} \cdot \frac{2}{11}, $$ where the first term is computed analogously to $P\{Y_i = 1\}$, and the second term comes from $\frac{\binom{1}{1}\binom{10}{1}}{\binom{11}{2}} = \frac{2}{11}$.
So, to sum it up, $EX = 5 \cdot \frac{3}{20} \cdot \frac{2}{11} + 8 \cdot \frac{3}{20} = \frac{147}{110}$, where $X$ is the white balls count in the final selection.