Finding least value of $m$ Let $B_1$,$B_2$ and $B_3$ be three boxes such that initially $B_1$ contains $2m$ balls( $m$ white balls and $m$ black balls), while $B_2$ and $B_3$ are empty. Now from $B_1$, $m$ balls are transferred to $B_2$ and remaining $m$ balls are transferred to $B_3$ (number of white and black balls in each being unknown). Then one ball is drawn from each box $B_2$ and $B_3$. Let $P(m)$ denotes the probability that these two drawn balls are of same color and it is found that $P(m)>\frac{1000}{2017}$ , then the minimum value of $m$ is?
 A: Let $B_{2,W}$ and $B_{2,B}$ be the random variables of the number of white and black balls in box 2, respectively.
Note that $B_{2,W}=i$ if and only if $B_{2,W}=m-i$, and that this also uniquely determines the number of white/black balls in $B_3$.
Then,
\begin{align}
P(B_{2,W}=i \cap B_{2,B}=m-i ) 
& = P(B_{2,W}=m-i \cap B_{2,B}=i )\\ 
% & = P(B_{3,W}=i \cap B_{3,B}=m-i )\\
% & = P(B_{3,W}=m-i \cap B_{3,B}=i )\\
& = \frac{{m \choose i}{m \choose m-i}}{{2m \choose m}}
% = \frac{{m \choose i}^2}{{2m \choose m}} 
\end{align}
Now let $A_{2,W}$ be the event: the ball from box 2 is white. And similarly define, $A_{3,W}$, $A_{2,B}$, $A_{3,B}$.
Using the law of total probability
\begin{align}
P(A_{2,W})= & P(A_{2,B})=P(A_{3,W})=P(A_{3,W})\\
= & \sum_{i=0}^m \frac{i}{m} \frac{{m \choose i}{m \choose m-i}}{{2m \choose m}} = \frac{1}{2}
\end{align}
because the RHS is the expected value (normalized by $m$) of a random variable having a hypergeometric distribution.
Then the probability that the two drawn balls from $B_2$ and $B_3$ have the same color (draws are assumed independent) is
\begin{align}
P(A_{2,W})P(A_{3,W}) + P(A_{2,B})P(A_{3,B}) = \frac{1}{2}
\end{align}
and therefore independent of $m$.
EDIT: As noted by antkam, this last passage is wrong because $A_{2,W}$ and $A_{3,W}$ are not independent.
A: From $B_1$, draw all $2m$ balls one by one and put them in a line.  Then form this mental picture:


*

*Imagine the first $m$ balls are in $B_2$.

*Imagine the second $m$ balls are in $B_3$.

*From $B_2$ you now need to draw a ball, but since all sequences are equally likely, you might as well just consider the first ball to be the drawn ball.

*From $B_3$ you now need to draw a ball, but since all sequences are equally likely, you might as well just consider the last ball to be the drawn ball.
So the event you want becomes: the first ball has the same color as the last ball.  This probability is clearly ${m-1 \over 2m-1}$.  So you solve:
$${m-1 \over 2m-1} > {1000 \over 2017} \Leftrightarrow 2017m-2017 > 2000m - 1000 \Leftrightarrow 17m > 1017 \Leftrightarrow m > {1017 \over 17} \approx 59.8     $$
So the minimum (integer) $m$ is 60.
