AMC 10 · 2024 · #17

Grade 8 probabilityalgebra

probability-basicsystematic-enumerationlinear-equations-one-var systematic-enumerationcaseworkconvert-to-algebra ↑ Prerequisites: probability-basicfraction-arithmetic

📏 Medium solution 💡 3 insights

Problem

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 $m + n$ ?

$\textbf{(A) } 10 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 13 \qquad \textbf{(E) } 14$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: In a best-of-three playoff, Team A hosts game 1 (win probability $\tfrac{2}{3}$) and Team B hosts games 2 and 3 (Team A's away win probability $p$). Game outcomes are independent and the playoff ends as soon as a team wins two games. Given that Team A's overall chance of winning the playoff is $\tfrac{1}{2}$, and $p$ can be written as $\tfrac{1}{2}(m - \sqrt{n})$ for positive integers $m, n$, find $m + n$.

Givens: $P(\text{A wins game 1}) = \tfrac{2}{3}$ (home), so $P(\text{B wins game 1}) = \tfrac{1}{3}$; $P(\text{A wins game 2}) = P(\text{A wins game 3}) = p$ (away), so $P(\text{B wins each}) = 1 - p$; Games are independent and the series stops the moment a team has $2$ wins; $P(\text{A wins the playoff}) = \tfrac{1}{2}$; $p = \tfrac{1}{2}(m - \sqrt{n})$ with $m, n$ positive integers; Answer choices: (A) $10$, (B) $11$, (C) $12$, (D) $13$, (E) $14$

Unknowns: The integers $m$ and $n$, and the sum $m + n$

Understand

Plan

Primary tool: #2 Make an Organized List

Secondary: #13 Convert to Algebra, #3 Eliminate Possibilities

A best-of-three has very few outcomes, so Tool #2 (Make an Organized List) is the right opening move: write down every game-by-game sequence in which A finishes with $2$ wins before B does, and stop. The list is short enough to enumerate by hand — exactly three patterns: $AA$, $ABA$, $BAA$. Once each sequence is given a probability in terms of $p$, Tool #13 (Convert to Algebra) takes over: add the three case probabilities, set the sum equal to $\tfrac{1}{2}$, and solve the resulting quadratic in $p$. The quadratic has two roots; Tool #3 (Eliminate Possibilities) discards the one that lies outside $[0,1]$, leaving a unique probability $p = \tfrac{1}{2}(4 - \sqrt{10})$ that matches the requested form.

Execute — Answer: E

#2 Make an Organized List 7.SP.C.8 Step 1

List every sequence of game results in which Team A wins the playoff.
The series stops as soon as either team reaches $2$ wins, so a sequence ends with A's second win.
Writing $A$ for an A-win and $B$ for a B-win, the only ways A can finish first are: $AA$ (sweep), $ABA$ (lose game 2 then win game 3), and $BAA$ (lose game 1 then win games 2 and 3).
These are mutually exclusive — they disagree on game 1 or game 2.

$$\text{Winning sequences for A: } \{AA,\; ABA,\; BAA\}$$

💡 Grade 7 "compound events with organized lists": when the sample space is small, write the outcomes down rather than try to count them in your head.

#2 Make an Organized List 7.SP.C.8 Step 2

Attach a probability to each sequence.
Game 1 is at A's home, so $P(A) = \tfrac{2}{3}$ and $P(B) = \tfrac{1}{3}$.
Games 2 and 3 are at B's home, so $P(A) = p$ and $P(B) = 1 - p$ each.
Independent games multiply.

$$P(AA) = \tfrac{2}{3} \cdot p = \tfrac{2p}{3}, \quad P(ABA) = \tfrac{2}{3}\cdot(1-p)\cdot p = \tfrac{2p(1-p)}{3}, \quad P(BAA) = \tfrac{1}{3}\cdot p \cdot p = \tfrac{p^2}{3}$$

💡 Each game outcome is independent, so the probability of a sequence is the product of the per-game probabilities — the standard Grade 7 compound-event move.

#13 Convert to Algebra 6.EE.B.7 Step 3

Add the three case probabilities and set the sum equal to $\tfrac{1}{2}$.
Then clear the denominator by multiplying both sides by $6$ and simplify into standard quadratic form.

$$\tfrac{2p}{3} + \tfrac{2p(1-p)}{3} + \tfrac{p^2}{3} = \tfrac{1}{2} \;\Rightarrow\; 4p + 4p - 4p^2 + 2p^2 = 3 \;\Rightarrow\; 2p^2 - 8p + 3 = 0$$

💡 Grade 6 "write and solve an equation that models the situation": translate the probability condition into one equation in the unknown $p$.

#13 Convert to Algebra 8.EE.A.2 Step 4

Solve the quadratic $2p^2 - 8p + 3 = 0$ with the quadratic formula.
The discriminant simplifies because $40 = 4 \cdot 10$ pulls a $2$ out of the radical.

$$p = \dfrac{8 \pm \sqrt{64 - 24}}{4} = \dfrac{8 \pm \sqrt{40}}{4} = \dfrac{8 \pm 2\sqrt{10}}{4} = \dfrac{4 \pm \sqrt{10}}{2}$$

💡 Grade 8 simplification of square roots: $\sqrt{40} = \sqrt{4}\cdot\sqrt{10} = 2\sqrt{10}$ keeps the radical small and exposes the requested form.

#3 Eliminate Possibilities 6.EE.B.8 Step 5

Eliminate the root that is not a valid probability.
Since $\sqrt{10} \approx 3.16$, the plus root gives $p \approx \tfrac{4 + 3.16}{2} \approx 3.58 > 1$, impossible for a probability.
The minus root gives $p \approx \tfrac{4 - 3.16}{2} \approx 0.42$, which sits in $[0,1]$.

$$p = \dfrac{4 - \sqrt{10}}{2} = \tfrac{1}{2}(4 - \sqrt{10}) \;\Rightarrow\; m = 4,\; n = 10$$

💡 Grade 6 "write an inequality to represent a constraint": $0 \le p \le 1$ is the constraint that throws out one root and locks in the other.

#13 Convert to Algebra 6.EE.A.2 Step 6

Read off $m + n$ from the matched form $p = \tfrac{1}{2}(m - \sqrt{n})$.

$$m + n = 4 + 10 = 14 \;\Rightarrow\; \textbf{(E)}$$

💡 Comparing two expressions in the same form to read off the values of $m$ and $n$ is the Grade 6 "letters stand for numbers" idea.

[1] #2 7.SP.C.8 List every sequence of game results in which Team A wins the playoff. The series

[2] #2 7.SP.C.8 Attach a probability to each sequence. Game 1 is at A's home, so $P(A) = \tfrac{

[3] #13 6.EE.B.7 Add the three case probabilities and set the sum equal to $\tfrac{1}{2}$. Then c

[4] #13 8.EE.A.2 Solve the quadratic $2p^2 - 8p + 3 = 0$ with the quadratic formula. The discrimi

[5] #3 6.EE.B.8 Eliminate the root that is not a valid probability. Since $\sqrt{10} \approx 3.1

[6] #13 6.EE.A.2 Read off $m + n$ from the matched form $p = \tfrac{1}{2}(m - \sqrt{n})$.

Review

Reasonableness: Plug $p = \tfrac{4 - \sqrt{10}}{2}$ back into the case sum. Numerically $p \approx 0.4189$, so $P(AA) \approx \tfrac{2}{3}(0.4189) \approx 0.2793$, $P(ABA) \approx \tfrac{2}{3}(0.4189)(0.5811) \approx 0.1623$, and $P(BAA) \approx \tfrac{1}{3}(0.4189)^2 \approx 0.0585$. Their sum is $0.2793 + 0.1623 + 0.0585 \approx 0.5001 \approx \tfrac{1}{2}$. The condition holds. Sign check: $\tfrac{2}{3} > \tfrac{1}{2}$, so Team A's home edge alone beats $50\%$, meaning the away probability $p$ has to be noticeably less than $\tfrac{1}{2}$ to drag the overall chance back down to $\tfrac{1}{2}$ — and $0.42$ fits that picture. Answer (E) $14$ is consistent.

Alternative: Tool #16 (Change Focus / Count the Complement): instead of summing A's three winning paths, sum B's three winning paths $BB$, $BAB$, $ABB$ and set their total equal to $\tfrac{1}{2}$. $P(BB) = \tfrac{1}{3}(1-p)$, $P(BAB) = \tfrac{1}{3}\cdot p \cdot (1-p)$, $P(ABB) = \tfrac{2}{3}(1-p)^2$. Adding and simplifying gives the same quadratic $2p^2 - 8p + 3 = 0$, hence the same $p = \tfrac{1}{2}(4 - \sqrt{10})$ and $m + n = 14$.

CCSS standards used (min grade 8)

7.SP.C.8 Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation (Listing the three mutually exclusive sequences ($AA$, $ABA$, $BAA$) in which Team A wins the playoff, then multiplying per-game probabilities to get the probability of each sequence.)
6.EE.B.7 Solve real-world and mathematical problems by writing and solving equations (Setting the total probability $\tfrac{2p}{3} + \tfrac{2p(1-p)}{3} + \tfrac{p^2}{3}$ equal to $\tfrac{1}{2}$ and clearing denominators to obtain $2p^2 - 8p + 3 = 0$.)
8.EE.A.2 Use square root and cube root symbols to represent solutions and evaluate square roots of small perfect squares (Simplifying $\sqrt{40} = 2\sqrt{10}$ inside the quadratic formula so that $p$ ends up in the form $\tfrac{4 \pm \sqrt{10}}{2}$ requested by the problem.)
6.EE.B.8 Write an inequality of the form $x > c$ or $x < c$ to represent a constraint or condition (Applying $0 \le p \le 1$ to reject the root $\tfrac{4 + \sqrt{10}}{2} > 1$ and keep $p = \tfrac{4 - \sqrt{10}}{2}$.)
6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers (Matching $\tfrac{4 - \sqrt{10}}{2}$ to $\tfrac{1}{2}(m - \sqrt{n})$ to read off $m = 4$ and $n = 10$.)

⭐ A short series has a short list of outcomes — write them all down, add their probabilities, and the unknown probability falls out of one quadratic equation. The $[0,1]$ rule for probability picks the right root.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 8)

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