Competition · AMC preparation · step 4 of 4

AMC 10 · 2017B · #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 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 13\frac{1}{3} and wrong with 23\frac{2}{3}.

P(right)=1/3, P(wrong)=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: 13\frac{1}{3}×13\frac{1}{3}×13\frac{1}{3}=127\frac{1}{27}.

P(exactly 3)=(1/3)³=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×(13\frac{1}{3}×13\frac{1}{3}×23\frac{2}{3})=627\frac{6}{27}.

P(exactly 2)=3×(1/3)²×2/3=3×2/27=6/27
5STEP 5

Add the two winning cases

Same denominator, so add numerators: 127\frac{1}{27}+627\frac{6}{27}=727\frac{7}{27}, the winning probability — choice (D).

P(win)=1/27+6/27=7/27 → (D)
Answer
7/27
The answer 727\frac{7}{27} is between 0 and 1, as every probability must be. It is a bit more than 14\frac{1}{4}, which feels right: winning by luck on hard random guesses should be possible but not likely, so an answer near 12\frac{1}{2} or tiny answers like 127\frac{1}{27} would both look wrong. The value 727\frac{7}{27}≈ 0.26 sits sensibly in the middle.
💡Key takeaway

Break "2 or more right" into exactly 3 right (127\frac{1}{27}) plus exactly 2 right (627\frac{6}{27}), add them to get 727\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

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