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Difference between revisions of "2010 AIME I Problems/Problem 4"

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== Problem ==
 
== Problem ==
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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>.
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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>.
  
 
== Solution ==
 
== Solution ==
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'''Case 1''': No heads.
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We perform [[casework]] based upon the number of heads that are flipped.  
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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
 
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<math>\frac {3}{28} = \frac {9}{784}</math>
 
  
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'''Case 2''': One head.
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*'''Case 1''': No heads.
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We can have either HTT, THT, or TTH. The first two happen to Jackie with the same <math>\frac {3}{28}</math> chance, but the third happens <math>\frac {4}{28}</math> 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.
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: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>
  
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Multiplying and adding up all 9 ways, we have a
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*'''Case 2''': One head.
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<math>\frac {4(3 \cdot 3) + 4(3 \cdot 4) + 1(4 \cdot 4)}{28^{2}} = \frac {100}{784}</math> overall chance for this case.
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: We can have either HTT, THT, or TTH. The first two happen to Jackie with the same <math>\frac {3}{28}</math> chance, but the third happens <math>\frac {4}{28}</math> 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.
  
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'''Case 3''': Two heads.
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:Multiplying and adding up all 9 ways, we have a <center><cmath>\frac {4(3 \cdot 3) + 4(3 \cdot 4) + 1(4 \cdot 4)}{28^{2}} = \frac {100}{784}</cmath></center>  
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With HHT <math>\frac {3}{28}</math>, HTH <math>\frac {4}{28}</math>, and THH <math>\frac {4}{28}</math> possible, we proceed as in Case 2, obtaining
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:overall chance for this case.
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<math>\frac {1(3 \cdot 3) + 4(3 \cdot 4) + 4(4 \cdot 4)}{28^{2}} = \frac {121}{784}</math>.
 
  
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'''Case 4''': Three heads.
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*'''Case 3''': Two heads.
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Similar to Case 1, we can only have HHH (<math>\frac {4}{28}</math> chance). Then in this case we get <math>\frac {16}{784}</math>
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:With HHT <math>\frac {3}{28}</math>, HTH <math>\frac {4}{28}</math>, and THH <math>\frac {4}{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>
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*'''Case 4''': Three heads.
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:Similar to Case 1, we can only have HHH, which has <math>\frac {4}{28}</math> chance. Then in this case we get <math>\frac {16}{784}</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>.
 
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>.
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{{AIME box|year=2010|num-b=3|num-a=5|n=I}}
 
{{AIME box|year=2010|num-b=3|num-a=5|n=I}}
  
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[[Category:Intermediate Algebra Problems]]
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[[Category:Intermediate Combinatorics Problems]]

Revision as of 12:25, 17 March 2010

Problem

Jackie and Phil have two fair coins and a third coin that comes up heads with probability $\frac47$. Jackie flips the three coins, and then Phil flips the three coins. Let $\frac {m}{n}$ be the probability that Jackie gets the same number of heads as Phil, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

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 $\frac {1}{2} \cdot \frac {1}{2} \cdot \frac {3}{7} = \frac {3}{28}$ Thus the probability for this to happen to both players is $\frac {3}{28} = \frac {9}{784}$
  • Case 2: One head.
We can have either HTT, THT, or TTH. The first two happen to Jackie with the same $\frac {3}{28}$ chance, but the third happens $\frac {4}{28}$ 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
\[\frac {4(3 \cdot 3) + 4(3 \cdot 4) + 1(4 \cdot 4)}{28^{2}} = \frac {100}{784}\]
overall chance for this case.
  • Case 3: Two heads.
With HHT $\frac {3}{28}$, HTH $\frac {4}{28}$, and THH $\frac {4}{28}$ possible, we proceed as in Case 2, obtaining
\[\frac {1(3 \cdot 3) + 4(3 \cdot 4) + 4(4 \cdot 4)}{28^{2}} = \frac {121}{784}.\]
  • Case 4: Three heads.
Similar to Case 1, we can only have HHH, which has $\frac {4}{28}$ chance. Then in this case we get $\frac {16}{784}$

Finally, we take the sum: $\frac {9 + 100 + 121 + 16}{784} = \frac {246}{784} = \frac {123}{392}$, so our answer is $123 + 392 = \fbox{515}$.

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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