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"

(Adding problem section)
m
 
(4 intermediate revisions by 3 users not shown)
Line 1: Line 1:
−
 
−
==Problem==
 
 
== Problem ==
 
== Problem ==
 
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==
−
=== Solution 1===
 
 
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> tails.
+
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>(3+4x)</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 product is <math>3+10x+11x^2+4x^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>1</math> head, and therefore <math>2</math> tails, 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 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>.
Line 20: Line 17:
 
<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 29: Line 26:
 
: 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.
 
: 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.
  
−
: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>
+
:Multiplying and adding up all 9 ways, we have a <cmath>\frac {4(3 \cdot 3) + 4(3 \cdot 4) + 1(4 \cdot 4)}{28^{2}} = \frac {100}{784}</cmath>
 +
 
 
:overall chance for this case.
 
:overall chance for this case.
  
 
*'''Case 3''': Two heads.
 
*'''Case 3''': Two heads.
 
: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
 
: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>
+
<cmath>\frac {1(3 \cdot 3) + 4(3 \cdot 4) + 4(4 \cdot 4)}{28^{2}} = \frac {121}{784}.</cmath>
  
 
*'''Case 4''': Three heads.
 
*'''Case 4''': Three heads.
Line 40: Line 38:
  
 
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 Also ==
 
== See Also ==

Latest revision as of 21:15, 27 August 2023

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 1

This can be solved quickly and easily with generating functions.

Let $x^n$ represent flipping $n$ heads.

The generating functions for these coins are $(1+x)$,$(1+x)$,and $(3+4x)$ 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 $10$ ways to get $1$ head, and therefore $2$ tails, here.)

The sum of the coefficients squared (total number of possible outcomes, squared because the event is occurring twice) is $(4 + 11 + 10 + 3)^2 = 28^2 = 784$ 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 $4^2 + 11^2 + 10^2 + 3^2=246$. The probability is then $\frac{4^2 + 11^2 + 10^2 + 3^2}{28^2} = \frac{246}{784} = \frac{123}{392}$. (Notice the relationship between the addends of the numerator here and the cases in the following solution.)

$123 + 392 = \boxed{515}$

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 $\frac {4}{28}$, HTH $\frac {4}{28}$, and THH $\frac {3}{28}$ 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_I_Problems/Problem_4&oldid=197540"