AMC10 2021 B

AMC10 2021 B · Q18

AMC10 2021 B · Q18. It mainly tests Probability (basic), Counting & probability misc.

A fair $6$-sided die is repeatedly rolled until an odd number appears. What is the probability that every even number appears at least once before the first occurrence of an odd number?

一个公平的 $6$ 面骰子反复掷直到出现奇数。每个偶数在第一次出现奇数之前至少出现一次的概率是多少？

(A) \frac{1}{120} \frac{1}{120}

(B) \frac{1}{32} \frac{1}{32}

(D) \frac{3}{20} \frac{3}{20}

(E) \frac{1}{6} \frac{1}{6}

Answer

Correct choice: (C)

正确答案：（C）

Solution

Since 3 out of 6 of the numbers are even, there is a $\frac36$ chance that the first number we choose is even. Since the number rolled first is irrelevant, we don't have to consider it. Therefore there are 2 even numbers out of the 5 choices left. There is a $\frac{2}{5}$ chance that the next number that is distinct from the first is even. There is a $\frac{1}{4}$ chance that the next number distinct from the first two is even. (There is only one even integer left. ) With all the even integers taken, the next integer rolled must be odd. $\frac{3}{6} \cdot \frac{2}{5} \cdot \frac{1}{4} = \frac{1}{20}$, so the answer is $\boxed{\textbf{(C) }\frac{1}{20}}.$

因为 $6$ 个数字中有 $3$ 个是偶数，所以第一次掷出偶数的概率是 $\frac{3}{6}$。由于第一次掷出的数字无关紧要，我们不必考虑它。因此剩余 $5$ 个选择中有 $2$ 个偶数。下一个与第一次不同的数字是偶数的概率是 $\frac{2}{5}$。下一个与前两个不同的数字是偶数的概率是 $\frac{1}{4}$。（只剩下一个偶数了。）所有偶数都出现后，下一个掷出的数字一定是奇数。 $\frac{3}{6} \cdot \frac{2}{5} \cdot \frac{1}{4} = \frac{1}{20}$，所以答案是 $\boxed{\textbf{(C) }\frac{1}{20}}$。

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