AMC10 2017 B

AMC10 2017 B · Q9

AMC10 2017 B · Q9. It mainly tests Combinations, Probability (basic).

A radio program has a quiz consisting of 3 multiple-choice questions, each with 3 choices. A contestant wins if he or she gets 2 or more of the questions right. The contestant answers randomly to each question. What is the probability of winning?

一个广播节目有一个测验，包括 3 个多选题，每个题有 3 个选项。参赛者答对 2 个或更多题则获胜。参赛者对每个题随机作答。获胜的概率是多少？

(A) \dfrac{1}{27} \dfrac{1}{27}

(B) \dfrac{1}{9} \dfrac{1}{9}

(D) \dfrac{7}{27} \dfrac{7}{27}

(E) \dfrac{1}{2} \dfrac{1}{2}

Answer

Correct choice: (D)

正确答案：（D）

Solution

The probability of getting all 3 questions right is (1/3)³ = 1/27. Because there are 3 ways to get 2 of the questions right and 1 wrong, the probability of getting exactly 2 right is 3(1/3)²(2/3) = 6/27. Therefore the probability of winning is 1/27 + 6/27 = 7/27.

答对全部 3 题的概率是 $\left(\frac{1}{3}\right)^3=\frac{1}{27}$。答对恰好 2 题（错 1 题）有 3 种方式，其概率是 $3\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)=\frac{6}{27}$。因此获胜概率是 $\frac{1}{27}+\frac{6}{27}=\frac{7}{27}$。

Topics