DIE AND COINS

This problem asks for the probability of an event given an observation. This makes it an excellent candidate for Bayes' Rule.

• Denote the probability of rolling 1, 2, 3, or 4 with the die as $P(1-4)$
• Denote the probability of rolling 5 or 6 with the die as $P(5-6)$
• Denote the probability of obtaining exactly one head as $P(1H)$
• We want to find the probability that the girl rolls a 1, 2, 3, or 4, given that we know that she obtained exactly one head: $P(1-4|1H)$

Bayes' theorem gives us the following relation:

• $P(1-4|1H) = \frac{P(1H|1-4)\cdot P(1-4)}{P(1H)}$

We evaluate $P(1H)$ by expanding it using conditional probabilities:

• $P(1H) = P(1H|1-4)\cdot P(1-4) + P(1H|5-6)\cdot P(5-6)$

We evaluate each of these expressions separately:

• Clearly $P(1-4) = \frac{2}{3}$ and $P(5-6) = \frac{1}{3}$
• If the girl rolls 1, 2, 3, or 4, then she tosses two coins: one biased $(P(H) = \frac{3}{4})$ and one fair $(P(H) = \frac{1}{2})$ . The probability of obtaining exactly one head is $P(1H|1-4) = \frac{3}{4}\cdot\frac{1}{2} + \frac{1}{4}\cdot\frac{1}{2} = \frac{1}{2}$
• If the girl rolls 5 or 6, then she tosses three coins: one is certain to give a head, and two with $P(H) = \frac{3}{4}$ . We will only see a total of one head if both of the uncertain coins come up tails. Therefore: $P(1H|5-6) = \left(\frac{1}{4}\right)^{2} = \frac{1}{16}$

This gives us all of the information we need to calculate the solution.

• $P(1-4|1H) = \frac{P(1H|1-4)\cdot P(1-4)}{P(1H)}$
• $P(1-4|1H) = \frac{\frac{1}{2}\cdot\frac{2}{3}}{\frac{1}{2}\cdot\frac{2}{3} + \frac{1}{16}\cdot\frac{1}{3}}$
• $P(1-4|1H) = \frac{16}{17}$

This means that $m=16$ and $n = 17$ , so $m+n = 33$ .