HMMT-Feb 2018 comb, Problem 5

A bag contains nine blue marbles, ten ugly marbles, and one special marble. Ryan picks marbles randomly from this bag with replacement until he draws the special marble. He notices that none of the marbles he drew were ugly. Given this information, what is the expected value of the number of total marbles he drew?

Solution: The probability of drawing $k$ marbles is the probability of drawing $k-1$ blue marbles and then the special marble, which is $p_{k}=\left(\frac{9}{20}\right)^{k-1} \times \frac{1}{20}$. The probability of drawing no ugly marbles is therefore $\sum_{k=1}^{\infty} p_{k}=\frac{1}{11}$. Then given that no ugly marbles were drawn, the probability that $k$ marbles were drawn is $11 p_{k}$. The expected number of marbles Ryan drew is \[ \sum_{k=1}^{\infty} k\left(11 p_{k}\right)=\frac{11}{20} \sum_{k=1}^{\infty} k\left(\frac{9}{20}\right)^{k-1}=\frac{11}{20} \times \frac{400}{121}=\frac{20}{11} \] (To compute the sum in the last step, let $S=\sum_{k=1}^{\infty} k\left(\frac{9}{20}\right)^{k-1}$ and note that $\frac{9}{20} S=S-\sum_{k=1}^{\infty}\left(\frac{9}{20}\right)^{k-1}=$ $\left.S-\frac{20}{11}\right)$. $\fbox{\frac{20}{11}}$.

Answer: \frac{20}{11}