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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$\underset{1}{\overset{5}{鈭�}}{X}_i+\underset{1}{\overset{8}{鈭�}}{Y}_i$

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

$N=\underset{i=1}{\overset{5}{鈭�}}{X}_i+\underset{i=1}{\overset{8}{鈭�}}{Y}_i$

$E\left(N\right)=\underset{i=1}{\overset{5}{鈭�}}E\left(X_i\right)+\underset{i=1}{\overset{8}{鈭�}}E\left(Y_i\right)$

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

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

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

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

The total number of chosen white balls is,

$E\left(N\right)=\underset{i=1}{\overset{5}{鈭�}}E\left(X_i\right)+\underset{i=1}{\overset{8}{鈭�}}E\left(Y_i\right)$

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