What is the probability that three babies are boys, given that at least one is a boy? I know this is a simple problem but I am arguing with a friend about its solution, so I want to show him an "official" proof!
Suppose that in any birth, the probability to have a boy is $48.5\%$.

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*If we have three persons expecting to deliver, what is the probability that at least one of them gives birth to a boy?

*If we know that at least one will give birth to a boy (suppose we have accurate ultra-sound results), what is the probability all three will have a boy?

For the first question, we calculate the probability of one NOT having a boy, which is $1-0.485 = 0.515$ and then the required probability of all three not having a boy is $0.515^3 = 0.1365$ so the probability that at least one will have a boy is $1-0.1365 = 0.8634 = 86.34\%$.
For the second question, since the three events are independent, the probability that all three will have a boy given that at least one will have a boy is equal to the probability that the other two will have a boy. Is it $0.485^2$? I am not sure about the second one.

Let $X\in\{0,1,2,3\}$ be the number of boys. Let $A$ be the event that $X=3$ and $B$ be the event that $X\ge1$. By definition, we have  ${\rm P}(A\cap B)={\rm P}(A\mid B){\rm P}(B)$ so $${\rm P}(X=3\mid X\ge1)=\frac{{\rm P}(X=3\cap X\ge1)}{{\rm P}(X\ge1)}=\frac{{\rm P}(X=3)}{{\rm P}(X\ge1)}.$$ You have calculated ${\rm P}(X\ge1)$ in the first part and evaluating ${\rm P}(X=3)$ is straightforward.
