Difference between revisions of "2010 AIME I Problems/Problem 4"
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Jackie and Phil have two fair coins and a third coin that comes up heads with [[probability]] <math>\frac47</math>. Jackie flips the three coins, and then Phil flips the three coins. Let <math>\frac {m}{n}</math> be the probability that Jackie gets the same number of heads as Phil, where <math>m</math> and <math>n</math> are [[relatively prime]] positive integers. Find <math>m + n</math>. | Jackie and Phil have two fair coins and a third coin that comes up heads with [[probability]] <math>\frac47</math>. Jackie flips the three coins, and then Phil flips the three coins. Let <math>\frac {m}{n}</math> be the probability that Jackie gets the same number of heads as Phil, where <math>m</math> and <math>n</math> are [[relatively prime]] positive integers. Find <math>m + n</math>. | ||
− | == Solution == | + | == Solution 1== |
+ | This can be solved quickly and easily with [[generating functions]]. | ||
+ | |||
+ | Let <math>x^n</math> represent flipping <math>n</math> tails. | ||
+ | |||
+ | The generating functions for these coins are <math>(1+x)</math>,<math>(1+x)</math>,and <math>(4+3x)</math> in order. | ||
+ | |||
+ | The product is <math>4+11x+10x^2+3x^3</math>. (<math>ax^n</math> means there are <math>a</math> ways to get <math>n</math> heads, eg there are <math>10</math> ways to get <math>2</math> heads, and therefore <math>1</math> tail, here.) | ||
+ | |||
+ | The sum of the coefficients squared (total number of possible outcomes, squared because the event is occurring twice) is <math>(4 + 11 + 10 + 3)^2 = 28^2 = 784</math> 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 <math>4^2 + 11^2 + 10^2 + 3^2=246</math>. | ||
+ | The probability is then <math> \frac{4^2 + 11^2 + 10^2 + 3^2}{28^2} = \frac{246}{784} = \frac{123}{392}</math>. | ||
+ | (Notice the relationship between the addends of the numerator here and the cases in the following solution.) | ||
+ | |||
+ | <math>123 + 392 = \boxed{515}</math> | ||
+ | |||
+ | == Solution 2== | ||
We perform [[casework]] based upon the number of heads that are flipped. | We perform [[casework]] based upon the number of heads that are flipped. | ||
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*'''Case 3''': Two heads. | *'''Case 3''': Two heads. | ||
− | :With HHT <math>\frac { | + | :With HHT <math>\frac {4}{28}</math>, HTH <math>\frac {4}{28}</math>, and THH <math>\frac {3}{28}</math> possible, we proceed as in Case 2, obtaining |
<center><cmath>\frac {1(3 \cdot 3) + 4(3 \cdot 4) + 4(4 \cdot 4)}{28^{2}} = \frac {121}{784}.</cmath></center> | <center><cmath>\frac {1(3 \cdot 3) + 4(3 \cdot 4) + 4(4 \cdot 4)}{28^{2}} = \frac {121}{784}.</cmath></center> | ||
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Finally, we take the sum: <math>\frac {9 + 100 + 121 + 16}{784} = \frac {246}{784} = \frac {123}{392}</math>, so our answer is <math>123 + 392 = \fbox{515}</math>. | Finally, we take the sum: <math>\frac {9 + 100 + 121 + 16}{784} = \frac {246}{784} = \frac {123}{392}</math>, so our answer is <math>123 + 392 = \fbox{515}</math>. | ||
− | == See | + | == See Also == |
{{AIME box|year=2010|num-b=3|num-a=5|n=I}} | {{AIME box|year=2010|num-b=3|num-a=5|n=I}} | ||
[[Category:Intermediate Combinatorics Problems]] | [[Category:Intermediate Combinatorics Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 23:01, 18 August 2019
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 tails.
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
- 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 |
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