Conditional probability "brain teaser"

Question

I'm reading A Practical Guide to Quantitative Finance Interviews by Xinfeng Zhou.

The book has the following two part question.

Part A: A company is holding a dinner for working mothers with at least one son. Ms. Jackson, a mother with two children, is invited. What is the probability that both children are boys?

Part B: Your new colleague, Ms. Parker is known to have two children. If you see her walking with one of her children and that child is a boy, what is the probability that both children are boys?

According to the book, in Part A, the answer is \begin{equation*} 1/3 = \frac{1/4}{3/4} = \frac{\mathbb{P}\{\text{two boys}\}}{\mathbb{P}\{\text{at least one boy out of two kids}\}} = \mathbb{P}\{\text{two boys} \, \mid \, \text{at least one out of two is a boy}\}, \end{equation*} while, in Part B, the answer is \begin{equation*} 1/2 = \mathbb{P}\{\text{second kid is a boy}\}. \end{equation*}

I would have thought the "right" answer --- in quotes since the wording is clearly (?) vague --- is $1/3$ in both cases. Am I missing something?

(Assume a kid is equally likely to be a boy or a girl...)

The best thing I can come up with is if we let $F_{1}$ and $F_{2}$ be coin flips taking values in the symbolic space $\{H,T\}$, then the solution to Part A is finding $\mathbb{P}\{(F_{1},F_{2}) = (H,H) \, \mid \, F_{1} \, \, \text{or} \, \, F_{2} = H\}$, while Part B is finding $\mathbb{P}\{F_{1} = H \, \mid \, F_{2} = H\}$. On the other hand, just because I "saw" the kid (or the coin flip), doesn't mean I know which kid (or which coin) it was so I don't see why "What is $\mathbb{P}\{F_{1} = H \, \mid \, F_{2} = H\}$?" should be the "right" interpretation of the question.

Answer 1

"Just because I 'saw' the kid (or coin flip) doesn't mean I know which kid (or which coin)..."

But in fact that kid (or coin) does become distinguishable, as it is the one you saw.

Answer 2

To explain the discrepancy between the two situations, in the first situation you can have either of the two children be boys. So call the children $A$ and $B$ then there are three possible situations, only $A$ is a boy, only $B$ is a boy or both $A$ and $B$ are boys.

In the second scenario, there are only 2 possible situations. If you say the boy you saw is $A$ then either $A$ is the only boy or $B$ is also a boy. The situation where $A$ is not a boy but $B$ is, is not a part of this condition.