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2017 AMC 10B Problem 9

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2017 AMC 10B Problem 9AMC 10 Step by Step
Name:                             Date: September 13, 2026
Basic Probability · difficulty 2 of 5 · about 3 min · position P1-10

A radio program has a quiz consisting of 33 multiple-choice questions, each with 33 choices. A contestant wins if he or she gets 22 or more of the questions right. The contestant answers randomly to each question. What is the probability of winning?

  1. A)

    127\frac{1}{27}

  2. B)

    19\frac{1}{9}

  3. C)

    29\frac{2}{9}

  4. D)

    727\frac{7}{27}

  5. E)

    12\frac{1}{2}

Answer

D

Key insight

Winning means exactly 2 right or all 3 right; count 3 ways for exactly two (each 1/27 times 2) plus 1 way for three, out of 27.

Solution

Each question is answered correctly with probability 13\frac13 and incorrectly with probability 23\frac23, independently. There are 33=273^3 = 27 equally likely answer patterns.

Exactly two right: choose which question is missed (33 ways), then that question has 22 wrong choices while the other two are forced. That is 3⋅2=63 \cdot 2 = 6 patterns.

All three right: 11 pattern.

Winning patterns: 6+1=76 + 1 = 7 out of 2727.

The answer is (D) 727\boxed{\textbf{(D)}\ \frac{7}{27}}.

Why this works

"Two or more" splits cleanly into "exactly two" and "exactly three," and each case is a product of independent probabilities times the number of positions where the failures can sit. Counting patterns out of 2727 equally likely outcomes keeps the bookkeeping concrete and avoids fraction arithmetic.

Alternative approach

Complement: losing means 00 or 11 right. Zero right has 23=82^3 = 8 patterns; exactly one right has 3⋅22=123 \cdot 2^2 = 12 patterns. So 27−20=727 - 20 = 7 winning patterns.

The trap

Forgetting the factor of 3 for which question is missed, giving (1/3)^2(2/3) + (1/3)^3 = 3/27.

Common mistakes

  • Forgetting the factor of 3 for which question is missed, giving (1/3)^2(2/3) + (1/3)^3 = 3/27.
  • Reading "2 or more" as "exactly 2" and answering 627=29\frac{6}{27} = \frac{2}{9}, choice (C).

Techniques

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