AMC8 2014

AMC8 2014 · Q18

AMC8 2014 · Q18. It mainly tests Combinations, Probability (basic).

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?

昨天市医院出生了四个孩子。假设每个孩子是男孩或女孩的概率相等。以下哪个结果最可能发生？

(A) All are boys All are boys

(B) All are girls All are girls

(D) 3 are the same gender and 1 is of the other gender 3 are the same gender and 1 is of the other gender

(E) All of these outcomes are equally likely All of these outcomes are equally likely

Answer

Correct choice: (D)

正确答案：（D）

Solution

We'll just start by breaking cases down. The probability of A occurring is $\left(\frac{1}{2}\right)^4 = \frac{1}{16}$. The probability of B occurring is $\left(\frac{1}{2}\right)^4 = \frac{1}{16}$. The probability of C occurring is $\dbinom{4}{2}\cdot \left(\frac{1}{2}\right)^4 = \frac{3}{8}$, 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 $\dbinom{4}{1}\cdot\left(\frac{1}{2}\right)^4 = \frac{1}{4}$ 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 $\frac{1}{4} \cdot 2 = \frac{1}{2}$. So out of the four fractions, D is the largest. So our answer is $\boxed{\text{(D) 3 are of one gender and 1 is of the other gender}}.$

我们从分情况讨论开始。事件A发生的概率是$\left(\frac{1}{2}\right)^4 = \frac{1}{16}$。事件B发生的概率也是$\left(\frac{1}{2}\right)^4 = \frac{1}{16}$。事件C发生的概率是$\dbinom{4}{2}\cdot \left(\frac{1}{2}\right)^4 = \frac{3}{8}$，因为我们需要从4个位置中选出2个是女孩。对于D，有两种可能情况，3个女孩和1个男孩，或3个男孩和1个女孩。第一种情况的概率是$\dbinom{4}{1}\cdot\left(\frac{1}{2}\right)^4 = \frac{1}{4}$，因为我们需要从4个位置中选出1个是男孩。然而第二种情况的概率相同，因为我们选择4个孩子中1个是女孩，所以总概率是$\frac{1}{4} \cdot 2 = \frac{1}{2}$。因此，在这四个概率中，D最大。所以我们的答案是$\boxed{\text{(D) 3个是同一性别，1个是另一性别}}.$

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