2004 AMC 10A Problem 10

Below is the professionally curated solution for Problem 10 of the 2004 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AMC 10A solutions, or check the answer key.

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Concepts:binomial probability casework

Difficulty rating: 1470

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?

$\dfrac{19}{128}$

$\dfrac{23}{128}$

$\dfrac{1}{4}$

$\dfrac{35}{128}$

$\dfrac{1}{2}$

Solution:

The two coins match when both show $0, 1, 2,$ or $3$ heads. Coin $A$ has weights $1, 3, 3, 1$ out of $8$ and coin $B$ has weights $1, 4, 6, 4, 1$ out of $16.$

The probability is $\dfrac{1\cdot 1 + 3\cdot 4 + 3\cdot 6 + 1\cdot 4}{8 \cdot 16} = \dfrac{35}{128}.$

Thus, the correct answer is D.

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