Problem of the Week (POTW)

Over the summer, a problem will be posted each week for students to work on and submit their solutions. These problems are curated by the HMMT Problems staff — the same people who write the HMMT November and February tests — and are intended to increase in difficulty every week, resetting with the start of a new month. After the submission deadline closes, the problem writers will review submissions and post one of their favorites alongside the next POTW. Discussion of the POTW through any form of media is closed until the submission deadline passes.

This Week's Problem (October 18 to October 22)

This week's problem will be released on Monday, October 18 at 12 PM EDT (UTC-0400).

Submission Form

The link to the submission form will be available once the problem has been released.

Schedule

The current POTW and past week’s solution will be posted on Monday, October 18 at 12 PM EDT (UTC-0400).
The deadline for submission is Friday, October 22 at 06 PM EDT (UTC-0400).

Last Week's Problem

We have two coins: a fair coin that comes up heads with probability $\frac12$ when flipped, and a biased coin that comes up heads with probability $p$. If we pick a coin at random and flip it, it comes up heads with probability $2p$. Given that a randomly picked coin came up heads, what is the probability that the coin is fair?

The answer to last week’s problem as $\boxed{3/4}$. Saksham S. writes
The probability that a coin comes up heads when randomly chosen and then flipped is $\frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2}\cdot p$, because either the fair coin is chosen or the biased coin is chosen. For the first case, there is a $1/2$ chance that the fair coin will get picked, and after getting picked, there is a $1/2$ chance it will come up heads. For the second case, there is a $1/2$ chance that the biased coin will get picked, and then a chance of $p$ that it will come up heads. We are given that this is $2p$, so \[\frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot p =2p \implies 1+2p=8p \implies 6p=1 \implies p=\frac{1}{6}.\]
To finish, Saksham uses the fact that the probability of event $A$ conditioning on event $B$ is $\mathbf{P}(A \text{ and }B)/\mathbf{P}(B)$, where $\mathbf{P}(\bullet)$ denotes the probability of an event. Therefore the answer is \[\frac{\mathbf{P}(\text{fair and heads})}{\mathbf{P}(\text{heads})} = \frac{1/4}{2p} = \frac{1/4}{1/3} = \frac{3}{4}.\]
Andrew L. takes a slightly different approach that doesn’t involve computing $p$ at all! The key realization is that \[\frac{\mathbf{P}(\text{fair and heads})}{\mathbf{P}(\text{heads})} = 1 - \frac{\mathbf{P}(\text{not fair and heads})}{\mathbf{P}(\text{heads})} = 1 - \frac{p/2}{2p} = \frac{3}{4}.\]

Archive

Past POTWs and their solutions can be found in the archive.