I first started with the assumption that probability of having boys and girls are same. I imagined I could get this using binomial probability distribution with number of trials=2, number of successes=2 and probability of success=0.5, which gives me probability of 1/4 for two boys. Here is where I got confused, since it doesn't matter in which order they had the children, only which gender they are. So considering all possible cases as BB, BG and GG and that they will be equiprobable, the correct answer for this case should be 1/3.
How do I get the probabilities of two boys from above if probability of having boys is say, p and having girls is (1-p)?
Best Answer
There are $4$ equally likely possibilities: BB, GG, BG, GB. Your first calculation is correct.
When the probability of a boy is $p$, the probability of two boys is $p^2$. We have two independent events.