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

Revision as of 21:12, 17 May 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 $m + n$ .

Solution

Solution 1

This can be solved quickly and easily with generating functions.

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

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

The product is $4+11x+10x^2+3x^3$ . ( $ax^n$ 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 $123 + 392 = \fbox{515}$ .

@@ Line 10: / Line 10: @@
 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>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>11</math> ways to get <math>2</math> heads, and therefore <math>1</math> tail, here.)
+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>11</math> ways to get <math>2</math> heads, and therefore <math>1</math> tail, here.)
 The sum of the coefficients squared is <math>784</math> and the sum of the squares of each coefficient is <math>246</math>.

2010 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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