AMC 8 2014 Problem 18

AMC8 2014 年第 18 题

概率★★★☆☆

Four children were born at City Hospital yesterday. Assume each child is equally likely to be a boy or a girl. Which of the following outcomes is most likely?

昨天城市医院出生了 4 个婴儿，假设每个婴儿是男孩还是女孩的概率是相等的。那么下面哪种结果是最有可能的？

A.All 4 are boys / 4 个都是男孩
B.All 4 are girls / 4 个都是女孩
C.2 are girls and 2 are boys / 2 个女孩和 2 个男孩
D.3 are of one gender and 1 is of the other gender / 3 个是同一性别，1 个是另一性别
E.All of these outcomes are equally likely / 以上结果可能性都相同

正确答案 · Correct Answer

解析 · Solution

Solution 1

We'll just start by breaking cases down. The probability of A occurring is

\displaystyle \left( \frac{1}{2} \right)^4 = \frac{1}{16}

. The probability of B occurring is

\displaystyle \left( \frac{1}{2} \right)^4 = \frac{1}{16}

. The probability of C occurring is

\displaystyle \dbinom{4}{2}\cdot \left( \frac{1}{2} \right)^4 = \frac{3}{8}

, because we need to choose 2 of the 4 slots to be girls. For D, there are two possible cases, 3 girls and 1 boy or 3 boys and 1 girl. The probability of the first case is

\displaystyle \dbinom{4}{1}\cdot\left( \frac{1}{2} \right)^4 = \frac{1}{4}

because we need to choose 1 of the 4 slots to be a boy. However, the second case has the same probability because we are choosing 1 of the 4 children to be a girl, so the total probability is

\displaystyle \frac{1}{4} \cdot 2 = \frac{1}{2}

. So out of the four fractions, D is the largest. So our answer is

\text{(D) 3 are of one gender and 1 is of the other gender}.

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