2004 AMC 10A Problem 10

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

"Two independent quantities are equal" is handled by conditioning on the common value: for each possible $k$, multiply the two ways of achieving it, then add over $k$. Because every flip sequence has the same probability, counting sequences and dividing by $128$ avoids fractions until the very end.

Note $\binom{4}{k} = \binom{4}{4-k}$, so the favorable count is $\sum_k \binom{3}{k}\binom{4}{4-k}$, which by the Vandermonde identity is the number of ways to choose $4$ objects from $3 + 4 = 7$: $\binom{7}{4} = 35$. Equivalently, "$A$'s heads equal $B$'s tails" among $7$ flips means exactly $4$ of the $7$ flips are heads-for-$A$-or-tails-for-$B$.

• Forgetting that the two coins' outcomes multiply, and adding the individual probabilities instead.
• Dividing by $2^3 + 2^4 = 24$ or by $2^{12}$ instead of $2^7$.