Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 1)
m (making solution sections look pretty)
Line 2: Line 2:
 
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 1==
+
==Solutions==
 +
=== Solution 1===
 
This can be solved quickly and easily with [[generating functions]].
 
This can be solved quickly and easily with [[generating functions]].
  
Line 17: Line 18:
 
<math>123 + 392 = \boxed{515}</math>
 
<math>123 + 392 = \boxed{515}</math>
  
−
== Solution 2==
+
=== 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.  
  
Line 41: Line 42:
  
  
−
== See also ==
+
== 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]]

Revision as of 15:52, 12 April 2012

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 $m + n$.

Solutions

Solution 1

This can be solved quickly and easily with generating functions.

Let $x^0 = 1$ represent heads and $x$ represent tails.

The generating functions for these coins are $(1+x)$,$(1+x)$,and $(4+3x)$ in order.

The product is $3+10x+11x^2+4x^3$. ($ax^n$ means there are $a$ ways to get $n$ heads, eg there are $11$ ways to get $2$ heads, and therefore $1$ tail, here.)

The sum of the coefficients squared is $784$ and the sum of the squares of each coefficient is $246$. The probability is then $\frac{246}{784} = \frac{123}{392}$.


$123 + 392 = \boxed{515}$

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 $\frac {1}{2} \cdot \frac {1}{2} \cdot \frac {3}{7} = \frac {3}{28}$ Thus the probability for this to happen to both players is $\left(\frac {3}{28}\right)^2 = \frac {9}{784}$
  • 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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_I_Problems/Problem_4&oldid=46264"