Easy conditional probability problem

Question

Suppose there are born as much girls as boys. Give the following probability:

What is the probability that a family with 2 children has both a boy and a girl as children, if the oldest child is a boy?

My attempt: Let $(B,G)$ be that the first child is a boy and the second one a girl and similarly for other ordered pairs.

Define $\Omega:= \{(B,B), (G,G), (B,G),(G,B)\}$

Then, the asked probability is:

$$\mathbb{P}(\{(J,M),(M,J)\} \mid\{(J,J), (M,J)\}) = \frac{\mathbb{P}(\{(M,J)\})}{\mathbb{P}(\{(J,J),(M,J)\})}$$

There is given that there are born as much girls as boys. Therefore, we can say that the probabilities are uniformly distributed, and hence, the conditional probability is:

$$\frac{1/4}{2/4}= 1/2$$

I ask this question because I'm not entirely sure why I can say that the probabilities are uniformly distributed. It seems given that $\mathbb{P}($a boy is born)$=$$\mathbb{P}($a girl is born) $= 1/2$

but how can I formally deduce that for example $\mathbb{P}(\{(M,J)\}) = 1/4$?

Answer 1

You are entirely right that the problem as stated says nothing about the children born in the same families being independent. For all we know this is an alien species where each male can only have offspring of one gender.

That being said, independence is most likely intended by the problem authors, and as long as you state that you make that assumption, you are in the clear.

Answer 2

The older child is a boy, so for this family to have a girl, the younger child must be a girl (and if the younger child is a girl, the family has a girl and a boy, so it's if and only if). The chances of the younger child being a girl is $\frac{1}{2}$, therefore the chances of there being a girl and a boy in the family is $\frac{1}{2}$.