2004 AMC 10A Problem 10

1. All outcomes. Coin $A$ has $2^3 = 8$ possible sequences and coin $B$ has $2^4 = 16$. Together there are $8 \times 16 = 128$ equally likely outcomes.
2. Good outcomes. Coin $A$ can only show $0$, $1$, $2$ or $3$ heads, so the common number of heads must be one of these. Handle each number separately.
3. For a given number of heads $k$, coin $A$ can get it in $\binom{3}{k}$ ways and coin $B$ in $\binom{4}{k}$ ways. Every way for $A$ can go with every way for $B$, so multiply.

1. Add the last column: $1 + 12 + 18 + 4 = 35$.
2. The probability is $\dfrac{35}{128}$.

The answer is $\boxed{\textbf{(D)}\ \frac{35}{128}}$.

To find the chance that two separate counts come out equal, go through each possible common value. For each one, multiply the number of ways each side can reach it, then add up. Counting outcomes out of $128$ keeps everything in whole numbers until the last step.

A slicker count. Coin $A$'s heads equal coin $B$'s heads exactly when (heads of $A$) plus (tails of $B$) equals $4$, because $B$'s heads and tails add to $4$. Think of all $7$ flips in a row and mark a flip if it is a head of $A$ or a tail of $B$. The good outcomes are the ones with exactly $4$ marked flips, and there are $\binom{7}{4} = 35$ of those.

All $2^3 \cdot 2^4 = 128$ combined flip sequences are equally likely, so count the sequences in which both coins show the same number of heads $k$. Coin $A$ shows $k$ heads in $\binom{3}{k}$ ways and coin $B$ in $\binom{4}{k}$ ways, and the choices are independent.

The favorable count is $1 + 12 + 18 + 4 = 35$, so the probability is $\dfrac{35}{128}$.

The answer is $\boxed{\textbf{(D)}\ \frac{35}{128}}$.