AMC 10 · 2017 · #9

Grade 7 probability

Archetype Overlap / Complement Counting └ Sub-archetype At-Least-One Complement: 1 − P(None)

probability-basiccaseworkcombinations-basic caseworkcomplementary-counting ↑ Prerequisites: probability-basic

📏 Medium solution 💡 2 insights

Problem

A quiz has 3 multiple-choice questions, each with 3 choices, and the contestant guesses every answer at random. A win means getting 2 or more of the 3 questions right. The question asks for the probability of winning.

Pick an answer.

(A)

$\frac{1}{27}$

(B)

$\frac{1}{9}$

(C)

$\frac{2}{9}$

(D)

$\frac{7}{27}$

(E)

$\frac{1}{2}$

AMC 10 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

"Two or more right" hides two separate situations, so Tool #7 (Identify Subproblems) leads: split winning into the case of exactly 3 right and the case of exactly 2 right, handle each on its own, then add. The exactly-2 case needs care because the one wrong answer can land on any of the three questions, so Tool #2 (Make a Systematic List) counts those 3 ways cleanly. Each single guess is right with probability $\frac{1}{3}$ , so each case is a product of these per-question chances. Tool #16 (Count the Complement) gives an independent check in review: instead of the two winning cases, add up the two losing cases and subtract from 1.

1STEP 1

Find the chance for one question

Three equally likely choices with one correct make a guess right with probability $\frac{1}{3}$ and wrong with $\frac{2}{3}$ .

P(right)=

\frac{1}{3}

, P(wrong)=

\frac{2}{3}

2STEP 2

Split winning into two cases

"2 or more" rules out 0 or 1 right, so it splits into exactly 3 right or exactly 2 right — two cases that never overlap.

P(win)=P(exactly 3)+P(exactly 2)

3STEP 3

Probability of all three right

All three must be right and guesses are independent, so multiply: $\frac{1}{3}$ × $\frac{1}{3}$ × $\frac{1}{3}$ = $\frac{1}{27}$ .

P(exactly 3)=(

\frac{1}{3}

)³=

\frac{1}{27}

4STEP 4

Probability of exactly two right

Exactly 2 right means one is wrong, and that wrong one is any of the 3 questions: 3×( $\frac{1}{3}$ × $\frac{1}{3}$ × $\frac{2}{3}$ )= $\frac{6}{27}$ .

P(exactly 2)=3×(

\frac{1}{3}

)²×

\frac{2}{3}

=3×

\frac{2}{27}

\frac{6}{27}

5STEP 5

Add the two winning cases

Same denominator, so add numerators: $\frac{1}{27}$ + $\frac{6}{27}$ = $\frac{7}{27}$ , the winning probability — choice (D).

P(win)=

\frac{1}{27}

\frac{6}{27}

\frac{7}{27}

→ (D)

Answer

\frac{7}{27}

The answer

\frac{7}{27}

is between 0 and 1, as every probability must be. It is a bit more than

\frac{1}{4}

, which feels right: winning by luck on hard random guesses should be possible but not likely, so an answer near

\frac{1}{2}

or tiny answers like

\frac{1}{27}

would both look wrong. The value

\frac{7}{27}

≈ 0.26 sits sensibly in the middle.

💡Key takeaway

Break "2 or more right" into exactly 3 right ( $\frac{1}{27}$ ) plus exactly 2 right ( $\frac{6}{27}$ ), add them to get $\frac{7}{27}$ , choice (D).

Find the chance for one question
Split winning into two cases
Probability of all three right
Probability of exactly two right
Add the two winning cases

Find the chance for one question

Split winning into two cases

Probability of all three right

Probability of exactly two right

Add the two winning cases

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