$3$ persons $A, B\ \&\ C$ has $10$ balls each. The colours of their balls are given in the table below. \begin{align} & Person & Red & & Blue & & Green \\ \hline & A & 6 & & 3 & & 1 \\ \hline & B & 2 & & 2 & & 6 \\ \hline & C & 2 & & 1 & & 7 \\ \hline \end{align} Now each one of them picks up a ball arbitrarily and gives to another person $D$. Their choices are independent of each other.
We are to find out the the expected number of blue balls $D$ will have.
My approach:
\begin{align} & P(A_{blue}) & = & &\frac{3}{10} & \tag{i} \\ & P(\overline{A_{blue}}) & = & &\frac{7}{10} & \tag{ii} \\ & P(B_{blue}) & = & &\frac{2}{10} & \tag{iii} \\ & P(\overline{B_{blue}}) & = & &\frac{8}{10} & \tag{iv} \\ & P(C_{blue}) & = & &\frac{1}{10} & \tag{v} \\ & P(\overline{C_{blue}}) & = & &\frac{9}{10} & \tag{vi} \\ \end{align} $P(A_{blue}) \equiv$ $A$'s probability of selecting $1$ blue ball.
Let by $P(D_n)$ we denote $D$'s probability of having $n$ blue balls.
So, \begin{align} & P(D_1) & = & & (i).(iv).(vi) + (ii).(iii).(vi) + (ii).(iv).(v) & & = & & \frac{398}{1000} \\ & P(D_2) & = & & (i).(iii).(vi) + (ii).(iii).(v) + (i).(iv).(v) & & = & & \frac{92}{1000} \\ & P(D_3) & = & & (i).(iii).(v) & & = & & \frac{6}{1000} \\ \end{align}
$\therefore E(D)\equiv$ Expected number of blue balls received by $D$ = $1.P(D_1) + 2.P(D_2) + 3.P(D_3) = \frac{6}{10}$, which is the exact same result as $P(A_{blue}) + P(B_{blue}) + P(C_{blue})$ (is this coincidental ?)
Is there any other elegant method for doing this ? (mine is too naive I suppose)
No, it is not coincidental.
Let $X_{A},X_{B},X_{C}$ denote the number of blue balls that $D$ receives from $A,B,C$ respectively. Then $D$ will receive $X:=X_{A}+X_{B}+X_{C}$ blue balls in total, and:
$$\mathbb{E}X=\mathbb{E}\left(X_{A}+X_{B}+X_{C}\right)=\mathbb{E}X_{A}+\mathbb{E}X_{B}+\mathbb{E}X_{C}=\frac{3}{10}+\frac{2}{10}+\frac{1}{10}=\frac{6}{10}$$
Here e.g. $X_{A}$ takes values in $\left\{ 0,1\right\} $ with $P\left\{ X_{A}=0\right\} =\frac{7}{10}$ and $P\left\{ X_{A}=1\right\} =\frac{3}{10}$ wich leads to $\mathbb{E}X_{A}=\frac{3}{10}=P\left\{ X_{A}=1\right\} =P\left(A_{blue}\right)$. Likewise $\mathbb{E}X_{B}=P\left(B_{blue}\right)$ and $\mathbb{E}X_{C}=P\left(C_{blue}\right)$.
Finding an expectation is often more easy then finding a distribution. Always try that route first.