AIME 2014 II · 第 6 题

Solution 1

The probability that he rolls a six twice when using the fair die is $\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}$. The probability that he rolls a six twice using the biased die is $\frac{2}{3}\times \frac{2}{3}=\frac{4}{9}=\frac{16}{36}$. Given that Charles rolled two sixes, we can see that it is $16$ times more likely that he chose the second die. Therefore the probability that he is using the fair die is $\frac{1}{17}$, and the probability that he is using the biased die is $\frac{16}{17}$. The probability of rolling a third six is

$$\frac{1}{17}\times \frac{1}{6} + \frac{16}{17} \times \frac{2}{3} = \frac{1}{102}+\frac{32}{51}=\frac{65}{102}$$ Therefore, our desired $p+q$ is $65+102= \boxed{167}$

Solution 2 (Official Solution)

This is an incredibly simple problem if one is familiar with conditional probability (SO BE FAMILIAR WITH IT)! The conditional probability that the third roll will be a six given that the first two rolls are sixes, is the conditional probability that Charles rolls three sixes given that his first two rolls are sixes. This is thus $\frac{\frac{1}{2}(\frac{2}{3})^3+\frac{1}{2}(\frac{1}{6})^3}{\frac{1}{2}(\frac{2}{3})^2+\frac{1}{2}(\frac{1}{6})^2}= \frac{65}{102}$. The answer is therefore $65+102= \boxed {167}$.

~th1nq3r

Note: I have just found out that this is also the official solution. So I did not STEAL it, but as a coincidence, have come across the EXACT SAME SOLUTION. LIKE EXACTLY THE SAME. I AM SLIGHTLY FRIGHTENED. :/