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2024 AMC 10A Problem 17

2024 AMC 10A Problem 17AMC 10 Step by Step

Name: Date: September 13, 2026

Basic Probability · difficulty 3 of 5 · about 4 min · position P16-20

Two teams are in a best-two-out-of-three playoff: the teams will play at most $3$ games, and the winner of the playoff is the first team to win $2$ games. The first game is played on Team A's home field, and the remaining games are played on Team B's home field. Team A has a $\frac{2}{3}$ chance of winning at home, and its probability of winning when playing away from home is $p$ . Outcomes of the games are independent. The probability that Team A wins the playoff is $\frac{1}{2}$ . Then $p$ can be written in the form $\frac{1}{2}(m - \sqrt{n})$ , where $m$ and $n$ are positive integers. What is $m+n$ ?

A)
$10$
B)
$11$
C)
$12$
D)
$13$
E)
$14$

Answer

Key insight

Summing the three winning scenarios, (2/3)p + (2/3)(1−p)p + (1/3)p² = 1/2 gives 2p² − 8p + 3 = 0, so p = (4 − √10)/2.

Solution

Game 1 is at home (A wins with probability $\tfrac23$ ); games 2 and 3 are away (A wins each with probability $p$ ). Team A wins the playoff in exactly three ways:

Win, win: $\tfrac23\cdot p$ .
Win, lose, win: $\tfrac23(1-p)p$ .
Lose, win, win: $\tfrac13\cdot p^2$ .

Adding and setting equal to $\tfrac12$ :
$\tfrac23 p + \tfrac23 p - \tfrac23 p^2 + \tfrac13 p^2 = \tfrac43 p - \tfrac13 p^2 = \tfrac12.$
Multiply by $6$ : $8p - 2p^2 = 3$ , i.e. $2p^2 - 8p + 3 = 0$ . The quadratic formula gives
$p = \frac{8 \pm \sqrt{64 - 24}}{4} = \frac{8 \pm 2\sqrt{10}}{4} = \frac{4 \pm \sqrt{10}}{2}.$
Since $p \le 1$ , take the minus sign: $p = \tfrac12(4 - \sqrt{10})$ , so $m = 4$ , $n = 10$ , and $m + n = 14$ .

The answer is $\boxed{\textbf{(E)}\ 14}$ .

Why this works

A short series has only a handful of winning paths; listing them and multiplying independent probabilities along each path is the whole setup. The distinct home and away probabilities just mean each game uses its own factor. The resulting equation in $p$ is quadratic, and the answer form $\tfrac12(m - \sqrt n)$ tells you to keep the root below $1$ .

The trap

Forgetting the win-lose-win scenario or using 2/3 for game 3; games 2 and 3 are both away, so both use p. Also reject the root above 1.

Common mistakes

Forgetting the win-lose-win scenario or using 2/3 for game 3; games 2 and 3 are both away, so both use p. Also reject the root above 1.
Sign or arithmetic errors when clearing fractions (e.g. getting $2p^2 - 8p - 3 = 0$ ).

Techniques

Split into exhaustive cases and handle each · Set up the equation/formula and compute; no special trick needed