女儿还是儿子

答案分别为：

$$\frac{1}{2},\ \frac{1}{3},\ \frac{1}{2}.$$

解释：

• 第 1 问与第 3 问都固定了“某个特定孩子”的性别，剩余那个孩子独立地以 $1/2$ 为女孩。
• 第 2 问只给出“至少一个男孩”的集合条件，样本空间是 $ {BB,BG,GB}$三种等可能，因此两男概率为$1/3$。

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Original Explanation

1/2, 1/3, 1/2

Solution

1. Since it is already known that younger one is a girl, we just need the probability that older one is also a girl. This is simply $1/2$ since both genders are equally likely as per the assumption in the question.
2. Without any information, there are 4 equally likely possibilities (BB, BG, GB, GG) for the genders of the two children. Since atleast-one of them is a boy, we can remove the GG case from our sample space. Now there are 3 possibilities, and all three are equally likely, hence the probability that both the children are girls is $1/3$
3. This is counter-intuitive. One might say that since the names (X, Y) were anonymous, this problem should have been same as the second problem. But it turns out that there is indeed some extra information, which changes the probability. Author has actually fixed a person (X), and X's gender is "G" from now on. Thus we only need to find the probability that Y's gender is "G".

It still confuses me how statment (2) and statement (3) are different statements, leading to different outcomes. Statement (1) is equivalent to statement (3).

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Conditional Probability

We can also calculate the second answer using conditional probability. We can denote:

$A$: the event that at least one child is a boy.

$B$: the event that both children are boys.

We want to find the probability of event $B$, given that $A$ has ocurred. Using conditional probability formula:

Here $P(B \cap A) = P(B) = 1/4$ (both $B$ and $A$ are true when $B$ is true)

$P(A)=3/4$ (amongst 4 equally likely cases, 3 are favourable)

overall, $P(B | A) = \dfrac{1/4}{3/4} = 1/3$