2017 AMC 10B Problem 9

Below is the professionally curated solution for Problem 9 of the 2017 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2017 AMC 10B solutions, or check the answer key.

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Concepts:binomial probabilitycasework

Difficulty rating: 1140

9.

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?

127\dfrac{1}{27}

19\dfrac{1}{9}

29\dfrac{2}{9}

727\dfrac{7}{27}

12\dfrac{1}{2}

Solution:

The probability that a contestant gets all 33 correct is 133=127.\frac 13^3 = \frac 1{27}. The probability that a contestant gets exactly 22 is 13223(32)=627.\frac 13^2\cdot \frac 23 \cdot \binom {3}{2} = \frac 6{27}. The combined probability is 627+127=727.\frac{6}{27} + \frac 1{27} = \frac {7}{27}.

Thus, the correct answer is D .

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