Difference between revisions of "2014 AIME II Problems/Problem 6"
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Revision as of 00:11, 21 May 2014
Problem
Charles has two six-sided die. One of the die is fair, and the other die is biased so that it comes up six with probability 
and each of the other five sides has probability 
. Charles chooses one of the two dice at random and rolls it three times. Given that the first two rolls are both sixes, the probability that the third roll will also be a six is 
, where 
and 
are relatively prime positive integers. Find 
.
Solution
The probability that he rolls a six twice when using the fair die is 
. The probability that he rolls a six twice using the biased die is 
. Given that Charles rolled two sixes, we can see that it is 
times more likely that he chose the second die. Therefore the probability that he is using the fair die is 
, and the probability that he is using the biased die is 
. The probability of rolling a third six is
![\[\frac{1}{17}\times \frac{1}{6} + \frac{16}{17} \times \frac{2}{3} = \frac{1}{102}+\frac{32}{51}=\frac{65}{102}\]](http://latex.artofproblemsolving.com/1/2/5/125239bc008928d0cf222ac7315615b46ad55d16.png)
Therefore, our desired is 
See also
| 2014 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
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