# Probability of choosing marbles from number of boxes

I'm having hard time figuring out the following question.

Suppose that for each k, there is a box containing k red marbles n-k green marbles. $(k=0,1,2,...,n)$ One box is chosen from these $n+1$ boxes, and then two marbles are selected at random without replacement from the chosen box. Find the probability that to marbles are red.

Initially I thought,

$$\binom{n+1}{1}\binom{n}{k}\binom{n-1}{k-1}$$ would be the number of ways to choose a box among $n+1$ boxes, and first and second red marble respectively. However, I got stuck, how do I even compute the sample space?

• You don't really need to do any counting. The probability you are trying to compute is just $\sum_k P(E_k) P(R_1 \mid E_k) P(R_2 \mid E_k \cap R_1)$, where $E_k$ is the event that the $k$-th box is chosen, $R_1$ is the event that the first marble selected is red, and $R_2$ is the event that the second marble is red. – Ben CW Oct 10 '16 at 3:08

Let $X$ be the box you sampled and $Y$ is the # of red marbles if you sample two marbles from $X$. Then $$\Pr(Y = 0 | X = k) = \frac{{n-k \choose 2}}{n \choose 2} = \frac{(n-k)(n-k-1)}{n(n-1)}$$ Therefore, \begin{align} \Pr(Y = 0) &= \sum_{k=0}^n \Pr(Y = 0 |X = k)\Pr(X = k) \\ &= \sum_{k=0}^n \frac{(n-k)(n-k-1)}{n(n-1)}\cdot \frac{1}{n+1} \\ &= \frac{1}{(n+1)n(n-1)}\cdot \sum_{k=0}^n (n-k)(n-k-1) \\ &= \frac{1}{(n+1)n(n-1)}\cdot \sum_{k=0}^n k(k-1) \end{align} Use $\sum_{k=0}^n k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{k=0}^n k = \frac{n(n+1)}{2}$ to simplify the formula above.
The probability that two marbles drawn from the same box selected without bias are both red will be given by the Law of Total Probability. Where $X$ is the count of red marbles drawn, and $K$ is the count of red marbles in the box selected, that is:
\begin{align}\mathsf P(X=2) ~=~& \sum_{k=1}^n \mathsf P(K=k)\mathsf P(X=2\mid K=k) \\[2ex]~=~& \frac 1{n+1}\sum_{k=2}^n \binom{k}{2}\Big/\binom{n}{2}\\[1ex]~\vdots~&\\[1ex] ~=~& \bbox[silver ,1pt]{\color{silver}{\frac 13}}\end{align}
And similarly for $\mathsf P(X=1), \mathsf P(X=0)$.