Problem: 昨天，四个孩子在城市医院出生。假设每个孩子是男孩或女孩的可能性是相等的。以下哪种结果最有可能？  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?．
Options:
aoVal: A
content: all $$4$$ are boys  4个都是男孩 
aoVal: B
content: all $$4$$ are girls  四个都是女孩 
aoVal: C
content: $$2$$ are girls and $$2$$ are boys  2个女孩和$$2$$个男孩 
aoVal: D
content: $$3$$ are of one gender and $$1$$ is of the other gender  3个是同一种性别，$$1$$个是另一种性别 
aoVal: E
content: all of these outcomes are equally likely  所有这些结果都有同样的可能性 
Answer: D
Analysis: 城市医院昨日诞下四名婴儿．假设每个婴儿各可能有一半几率为男孩或女孩．则下列哪个事件最有可能？  A．$$4$$个男婴  B．$$4$$个女婴  C．$$2$$男婴$$2$$女婴  D．$$3$$个为一种性别，另一个为异性  E．以上事件皆等可能  我们先把例子分析一下．$$A$$发生的概率为$${{\left( \frac{1}{2} \right)}^{4}}=\frac{1}{16}$$，$$B$$发生的概率为$${{\left( \frac{1}{2} \right)}^{4}}=\frac{1}{16}$$．$$C$$发生的概率为$$\left( \begin{matrix}4    2   \end{matrix} \right)\cdot {{\left( \frac{1}{2} \right)}^{4}}=\frac{3}{8}$$．因为我们需要从$$4$$个孩子中选择$$2$$个成为女孩．  $$D$$方面，有两种可能的情况，即$$3$$个女孩和$$1$$个男孩或$$3$$个男孩和$$1$$个女孩．第一种情况的概率就$$\left( \begin{matrix}4    1   \end{matrix} \right)\cdot {{\left( \frac{1}{2} \right)}^{4}}=\frac{1}{4}$$．因为我们需要从四个孩子中选择一个成为男孩．但是，第二种情况也有同样的可能性，因为我们在$$4$$个孩子中选择了$$1$$个是女孩，所以总的可能性是$$\frac{1}{4}\cdot 2=\frac{1}{2}$$．  所以在四个分数中，$$D$$就最大的．所以我们的答案是$$3$$ are of one gender and $$1$$ is of the other gender．  故选$$\text{D}$$．  可能性列在帕斯卡三角形的第四行，最左边的$$1$$是所有男孩的可能性，最右边的$$1$$是所有女孩的可能性．由于 pascal 的三角形第四行为$$1$$，$$4$$，$$6$$，$$4$$，$$1$$，$$6$$，均为每个性别的两个孩子的可能性，因此共有$$8$$个可能性，其中一个性别的三个孩子和另一个性别的一个孩子．由于总共有$$2^{4}=16$$个儿童性别的可能性．  故选$$\text{D}$$．  We\textquotesingle 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 $$\left( \begin{matrix}4    2   \end{matrix} \right)\cdot {{\left( \frac{1}{2} \right)}^{4}}=\frac{3}{8}$$, because we need to choose $$2$$ of the $$4$$ children 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 $$\left( \frac{4}{1} \right)\cdot {{\left( \frac{1}{2} \right)}^{4}}=\frac{1}{4}$$ because we need to choose $$1$$ of the $$4$$ children 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 $$\left( \text{D} \right)$$ $$3$$ of one gender and $$1$$ of the other.  The possibilities are listed out in the fourth row of Pascal\textquotesingle s triangle, with the leftmost $$1$$ being the possibility of all boys and the rightmost $$1$$ being the possibility of all girls. Since the fourth row of Pascal\textquotesingle s Triangle goes $$1$$, $$4$$, $$6$$, $$4$$, $$1$$ and $$6$$ are all the possibilities of two children from each gender, there are a total of $$8$$ possibilities of three children from one gender and one from the other. Since there are a total of $${{2}^{4}}=16$$ total possibilities for the gender of the children, $$\text{D}$$ has the highest probability. 
Knowledge points: 美国AMC8->Knowledge Point->Counting, Probability and Statistics->Statistical Distribution
课内体系->知识点->统计与概率->概率->可能性的大小
Source: 2014年全美数学竞赛（AMC）竞赛