Probability of getting a white a ball after $4$ balls are drawn A bag contain $6$ Red and $4$ White balls. $4$ balls are drawn one by one without replacement and were found to
be at least $2$ white.What is the probability that next draw of a ball from this bag will give a white ball.?
My try:
As there were at least $2$ white balls, hence I made three cases.
$2W$ and $2R$ or $3W$ and $1R$ or $4W$
Now
Required probability=$\frac{C(4,2)C(6,2)}{C(10,4)}\left(\frac{2}{6}\right)+\frac{C(4,3)C(6,1)}{C(10,4)}\left(\frac{1}{6}\right)+0$
But it doesn't give right answer which is $\frac{34}{115}$. What is mistake in my method? How to get correct answer?
 A: I think you just forgot a "denominator".
Let $N=4$ be the number of white balls, $M=6$ the number of red balls and $d=4$ the number of ball to sample at the first round. Then, you want to calculate the probability that A="your $(d+1)$-th sample is a white ball" given that B="the previous $d$ samples gave at least two white balls", i.e., $P(A|B)$.
Thus we have $P(A|B)= \frac{P(A \cap B )}{ P(B)}$.
Now, the probability of selecting exactly $k$ white balls follows a hypergeometric distribution, so given the specific choice of $N,M$ and $d$, we get
$$
P(B) = \sum_{k=2}^{d} \frac{ {N \choose k} {M \choose d-k} }{ {N+M \choose d} }
$$
For the event $A \cap B$, note that we want to sample $(d+1)$ balls in total and
$$
P(A\cap B) = \sum_{k=2}^{d}  \frac{N-k}{N+M-d} \frac{ {N \choose k} {M \choose d-k} }{ {N+M \choose d} }.
$$
In fact the first term is the conditional probability of selecting a white ball given that in the first $d$ samples we got $k$, and the second term is the probability of selecting $k$ white balls in the first $d$ samples.
