Probability that the second ball is red n balls, each equally likely to be red or black, are added to a box containing one red ball. Given that the first ball withdrawn from the box is red, what's the probability that the second ball withdrawn from the box is red?
I figured that the answer is (n+3)/(2n+4), but I don't know how to prove that.
 A: Let $A_1$ denote the event that the first chosen ball is red; and $A_2$ the event  that the first chosen ball is red. I assume no replacement. 
Let $B_k$ (for $k \in \{0,...n\}$) denote the event that $k$ of the $n$ balls added to the box are red.
Then 
$$P(A_2| A_1) = \dfrac{P(A_1 \cap A_2)}{P(A_1)}.$$
Now
$$P(A_1 \cap A_2)  = \sum_{k=0}^n P(A_1 \cap A_2|B_k)P(B_k) = \sum_{k=1}^n P(A_1 \cap A_2|B_k)P(B_k),$$
the second inequality being true because you can't pick two red balls if there is only one in the box. 
Now 
$$P(A_1 \cap A_2|B_k) = \dfrac{k(k+1)}{n(n+1)}, $$
and 
$$P(B_k) = {n \choose k}(\dfrac{1}{2})^n.$$
So 
$$P(A_1 \cap A_2) = \sum_{k=1}^n \dfrac{k(k+1)}{n(n+1)} {n \choose k}(\dfrac{1}{2})^n.$$
By similar reasoning,
$$P(A_1) = \sum_{k=0}^n \dfrac{k+1}{n+1} {n \choose k}(\dfrac{1}{2})^n.$$
It shouldn't be too hard to calculate these sums explicitly. Then you just need to take the quotient. 
I hope this helps,
Frank.
A: The number $Y$ of added red balls has binomial distribution, parameters $n$ and $\frac{1}{2}$. We assume that the withdrawing is done without replacement. 
Let $A$ be the event the first ball withdrawn is red, and let $B$ be the event the second ball withdrawn is red. We want $\Pr(B|A)$, which is $\Pr(A\cap B)/\Pr(A)$.
Now compute. The probability of $A$ is 
$$\sum_0^n \frac{k+1}{n+1}\binom{n}{k}\frac{1}{2^n}.\tag{1}$$
The probability of $A\cap B$ is
$$\sum_0^n \frac{(k+1)(k)}{(n+1)(n)}\binom{n}{k}\frac{1}{2^n}.\tag{2}$$
Divide. 
That's the answer, but presumably we want to simplify. To calculate $\Pr(A)$, we want to calculate $\sum_0^n (k+1)\binom{n}{k}$. 
Recall that $(1+x)^n=\sum_0^n \binom{n}{k}x^k$ and therefore $x(1+x)^n=\sum_0^n \binom{n}{k}x^{k+1}$. Differentiate, and set $x=1$. We find that $\sum_0^n(k+1)\binom{n}{k}=2^n+n2^{n-1}=(n+2)2^{n-1}$.
Thus $\Pr(A)=\frac{n+2}{2(n+1)}$.
A similar but somewhat more tedious argument gets us a closed form for $\Pr(A\cap B)$. Differentiate twice instead of once. 
A: Hint:  You can imagine that the ball that started out in the box is green, and it is the first ball drawn.  Then what is the chance the next ball drawn is red?
