Suppose an urn contains $r$ red balls and $b$ blue balls. Suppose $n$ balls are drawn sequentially without replacement. Find the probability that $k$ of the $n$ balls are blue and that the first one is blue.
My try:
Let $A$ be event first ball is blue and $B$ be event $k$ of $n$ balls are blue. I want to find $P(A \cap B)$. In total we have ${r+b \choose n }$ outcomes in sample space. for the number of ways $A \cap B$ occurs we have first one is blue, and so we have $b-1$ blue balls from there we pick $k$ of them so ${b – 1 \choose k}$ and for red balls we have ${r \choose n – k}$. Therefore
$$ P(A \cap B) = \frac{ b {b – 1 \choose k} {r \choose n – k} }{{r+b \choose n } }. $$
Is this correct?
Best Answer
$$P(A\cap B)=P(B)P(A|B)$$ $$P(B)=\frac{\binom bk\binom r{n-k}}{\binom{r+b}n}$$ $$P(A|B)=\frac kn$$ Alternatively $$P(A\cap B)=P(A)P(B|A)$$ $$P(A)=\frac b{r+b}$$ $$P(B|A)=\frac{\binom{b-1}{k-1}\binom r{n-k}}{\binom{r+b-1}{n-1}}$$