AMC10 2004 10A, Problem 10

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?

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: $\frac{1+12+18+4}{128}=\frac{35}{128}\Rightarrow\fbox{\frac{35}{128}}$

Answer: \frac{35}{128}