Difference between revisions of "2010 AIME I Problems/Problem 4"
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This can be solved quickly and easily with [[generating functions]]. | This can be solved quickly and easily with [[generating functions]]. | ||
| − | Let <math>x^n</math> represent flipping <math>n</math> | + | Let <math>x^n</math> represent flipping <math>n</math> heads. |
The generating functions for these coins are <math>(1+x)</math>,<math>(1+x)</math>,and <math>(4+3x)</math> in order. | The generating functions for these coins are <math>(1+x)</math>,<math>(1+x)</math>,and <math>(4+3x)</math> in order. | ||
Revision as of 16:19, 16 July 2022
Contents
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 1
This can be solved quickly and easily with generating functions.
Let 
represent flipping heads.
The generating functions for these coins are 
,,and 
in order.
The product is 
. (
means there are 
ways to get heads, eg there are 
ways to get 
heads, and therefore 
tail, here.)
The sum of the coefficients squared (total number of possible outcomes, squared because the event is occurring twice) is 
and the sum of the squares of each coefficient (the sum of the number of ways that each coefficient can be chosen by the two people) is 
.
The probability is then 
.
(Notice the relationship between the addends of the numerator here and the cases in the following solution.)
Solution 2
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 
Thus the probability for this to happen to both players is 
- 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.
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
![\[\frac {4(3 \cdot 3) + 4(3 \cdot 4) + 1(4 \cdot 4)}{28^{2}} = \frac {100}{784}\]](//latex.artofproblemsolving.com/d/1/2/d1210732e708c01571d7cde20e973bdaeea18ae1.png)
- overall chance for this case.
![\[\frac {4(3 \cdot 3) + 4(3 \cdot 4) + 1(4 \cdot 4)}{28^{2}} = \frac {100}{784}\]](http://latex.artofproblemsolving.com/d/1/2/d1210732e708c01571d7cde20e973bdaeea18ae1.png)
- Case 3: Two heads.
- With HHT , HTH , and THH possible, we proceed as in Case 2, obtaining
![\[\frac {1(3 \cdot 3) + 4(3 \cdot 4) + 4(4 \cdot 4)}{28^{2}} = \frac {121}{784}.\]](http://latex.artofproblemsolving.com/7/4/9/7499bc0145c48d7afd1373e83db7cf746fc18412.png)
- 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
