Difference between revisions of "2010 AIME I Problems/Problem 4"
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*'''Case 1''': No heads. | *'''Case 1''': No heads. | ||
− | :The only possibility is TTT (the third coin being the unfair coin). The probability for this to happen to Jackie is <math>\frac {1}{2} \cdot \frac {1}{2} \cdot \frac {3}{7} = \frac {3}{28}</math> Thus the probability for this to happen to both players is <math>\frac {3}{28} = \frac {9}{784}</math> | + | :The only possibility is TTT (the third coin being the unfair coin). The probability for this to happen to Jackie is <math>\frac {1}{2} \cdot \frac {1}{2} \cdot \frac {3}{7} = \frac {3}{28}</math> Thus the probability for this to happen to both players is <math>\left(\frac {3}{28})\right^2 = \frac {9}{784}</math> |
*'''Case 2''': One head. | *'''Case 2''': One head. |
Revision as of 00:41, 24 February 2011
Problem
Jackie and Phil have two fair coins and a third coin that comes up heads with probability . Jackie flips the three coins, and then Phil flips the three coins. Let be the probability that Jackie gets the same number of heads as Phil, where and are relatively prime positive integers. Find .
Solution
We perform casework based upon the number of heads that are flipped.
- Case 1: No heads.
- The only possibility is TTT (the third coin being the unfair coin). The probability for this to happen to Jackie is Thus the probability for this to happen to both players is $\left(\frac {3}{28})\right^2 = \frac {9}{784}$ (Error compiling LaTeX. )
- Case 2: One head.
- We can have either HTT, THT, or TTH. The first two happen to Jackie with the same chance, but the third happens of the time, since the unfair coin is heads instead of tails. With 3 possibilities for Jackie and 3 for Phil, there are a total of 9 ways for them both to have 1 head.
- Multiplying and adding up all 9 ways, we have a
- overall chance for this case.
- Case 3: Two heads.
- With HHT , HTH , and THH possible, we proceed as in Case 2, obtaining
- Case 4: Three heads.
- Similar to Case 1, we can only have HHH, which has chance. Then in this case we get
Finally, we take the sum: , so our answer is .
See also
2010 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |