problem with the expected value

Question

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?

Answer 1

When you transfer two balls from Urn $1$ to Urn $2$, they can either be $WW$, $WB$, or $BB$. Then we have $$ P(WW\text{ added to Urn $2$})=\frac{\binom{5}{2}}{\binom{11}{2}}=\frac{2}{11}\\ P(WB\text{ added to Urn $2$})=\frac{\binom{5}{1}\binom{6}{1}}{\binom{11}{2}}=\frac{6}{11}\\ P(BB\text{ added to Urn $2$})=\frac{\binom{6}{2}}{\binom{11}{2}}=\frac{3}{11}. $$

Let $X_{WW}$ be the random variable that counts the number of white balls drawn from Urn $2$ after $WW$ is transferred from Urn $1$. Then $$ \begin{align*} E(X_{WW})&=0\cdot P(X_{WW}=0)+1\cdot P(X_{WW}=1)+2\cdot P(X_{WW}=2)+3\cdot P(X_{WW}=3)\\ &=P(X_{WW}=1)+2\cdot P(X_{WW}=2)+3\cdot P(X_{WW}=3)\\ &=\frac{\binom{10}{1}\binom{10}{2}+2\binom{10}{2}\binom{10}{1}+3\binom{10}{3}\binom{10}{0}}{\binom{20}{3}}\\ &=\frac 64. \end{align*} $$

Similarly we have $$ \begin{align*} E(X_{WB})&=0\cdot P(X_{WB}=0)+1\cdot P(X_{WB}=1)+2\cdot P(X_{WB}=2)+3\cdot P(X_{WB}=3)\\ &=P(X_{WB}=1)+2\cdot P(X_{WB}=2)+3\cdot P(X_{WB}=3)\\ &=\frac{\binom{9}{1}\binom{11}{2}+2\binom{9}{2}\binom{11}{1}+3\binom{9}{3}\binom{11}{0}}{\binom{20}{3}}\\ &=\frac{27}{20} \end{align*} $$ and $$ \begin{align*} E(X_{BB})&=0\cdot P(X_{BB}=0)+1\cdot P(X_{BB}=1)+2\cdot P(X_{BB}=2)+3\cdot P(X_{BB}=3)\\ &=P(X_{BB}=1)+2\cdot P(X_{BB}=2)+3\cdot P(X_{BB}=3)\\ &=\frac{\binom{8}{1}\binom{12}{2}+2\binom{8}{2}\binom{12}{1}+3\binom{8}{3}\binom{12}{0}}{\binom{20}{3}}\\ &=\frac 65. \end{align*} $$ Finally if we let $X$ count the number of white balls drawn from Urn $2$ under any condition then $$ E(X)=\frac{2}{11}\cdot\frac{6}{4}+\frac{6}{11}\cdot\frac{27}{20}+\frac{3}{11}\cdot\frac{6}{5}=\frac{147}{110} $$

Answer 2

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.