2004 AMC 10A Problem 10

P1-10~3 minPrint

Basic Probability Basic Counting Split into exhaustive cases and handle each

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?

Show solution without answering

Key insight

Go through each possible number of heads, 0 to 3. Multiply the ways coin A and coin B can each get that many, then add: 35 out of 128.

Solution

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.
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.
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.

heads $k$	ways for $A$	ways for $B$	ways for both
0	1	1	1
1	3	4	12
2	3	6	18
3	1	4	4

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

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

Why this works

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.

Alternative approach

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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.

$k$	$\binom{3}{k}$	$\binom{4}{k}$	product
0	1	1	1
1	3	4	12
2	3	6	18
3	1	4	4

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

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

Review in the AoPS books

Introduction to Counting & Probability, Chapter 7: Introduction to Probability
Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)
Introduction to Counting & Probability, Chapter 8: Basic Probability Techniques