# 2004 AMC 10A Problems/Problem 10

## Problem

Coin $A$ is flipped three times and coin $B$ is flipped four times. What is the probability that the number of heads obtained from flipping the two fair coins is the same? $\mathrm{(A) \ } \frac{29}{128} \qquad \mathrm{(B) \ } \frac{23}{128} \qquad \mathrm{(C) \ } \frac14 \qquad \mathrm{(D) \ } \frac{35}{128} \qquad \mathrm{(E) \ } \frac12$

## Solution

There are $4$ ways that the same number of heads will be obtained; $0$, $1$, $2$, or $3$ heads.

The probability of both getting $0$ heads is $\left(\frac12\right)^3{3\choose0}\left(\frac12\right)^4{4\choose0}=\frac1{128}$

The probability of both getting $1$ head is $\left(\frac12\right)^3{3\choose1}\left(\frac12\right)^4{4\choose1}=\frac{12}{128}$

The probability of both getting $2$ heads is $\left(\frac12\right)^3{3\choose2}\left(\frac12\right)^4{4\choose2}=\frac{18}{128}$

The probability of both getting $3$ heads is $\left(\frac12\right)^3{3\choose3}\left(\frac12\right)^4{4\choose3}=\frac{4}{128}$

Therefore, the probabiliy of flipping the same number of heads is: $\frac{1+12+18+4}{128}=\frac{35}{128}\Rightarrow\boxed{\mathrm{(D)}\ \frac{35}{128}}$

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