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.

Hint: LetXi = $1$if the i th white ball initially in urn $1$is one of the three selected, and let Xi = $0$otherwise. Similarly, let Yi = $1$if the i the 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$\sum _1^5{X}_i+\sum _1^8{Y}_i$

Short answer

The total number of chosen white balls is,

$N=\sum _{i=1}^5{X}_i+\sum _{i=1}^8{Y}_i$
E(N) =$\frac{147}{110}$

Step-by-step solution

Given Information

No.of balls in urn $1$are White balls and $6$Black ball. No. of balls in urn $2$are $8$White balls and $10$Black balls. LetXi = $1$if the i th white ball initially in urn $1$is one of the three selected, and let Xi = $0$otherwise. Similarly,letYi =$1$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$\sum _1^5{X}_i+\sum _1^8{Y}_i$

Explanation

Define indicator random variables Xi that are equal to $1$if and only if i th white ball from the urn 1 has been chosen in the trio, i=$1$, ..., $5$. Similarly, define indicator random variables Yi that are equal to 1 if and only if i th white ball from the urn $2$has been chosen in the trio, i=$1$, ..., $8$. The total number of chosen white balls is

$N=\sum _{i=1}^5{X}_i+\sum _{i=1}^8{Y}_i$

and using the linearity of the expectation, we have that

$E\left(N\right)=\sum _{i=1}^{5}E\left(X_i\right)+\sum _{i=1}^{8}E\left(Y_i\right)$

Explanation

The probability that a certain White ball from urn $1$is contained in the trio is equal to

$E\left(X_i\right)=P(X_i=1)=\frac{2}{11}\cdot \frac{3}{20}$

The probability that a certain White ball from urn $2$is contained in the trio is equal to

$E\left(Y_i\right)=P(Y_i=1)=\frac{3}{20}$

Therefore

$E\left(N\right)=5\cdot \frac{2}{11}\cdot \frac{3}{20}+8\cdot \frac{3}{20}=\frac{147}{110}$

Final Answer

The total number of chosen white balls is,

$E\left(N\right)=\sum _{i=1}^{5}E\left(X_i\right)+\sum _{i=1}^{8}E\left(Y_i\right)$

E(N) =$\frac{147}{110}$