So, it goes like this. We choose families at random from the families with two children (excluding twins). This is in a world where boys and girls are equally likely, and can be born equally on any day of the week.
We are told that family A has at least one boy. Then the probability that both are boys is 33%. (1/3)
We are told that family B has at least one boy born on a Tuesday. Then the probability that both are boys is 48%. (13/27)
@chomalungma, @waterlego - let me have a stab at painting the picture:
Imagine a big hall with all the children (in a statistically representative sample) paired up with their sibling. We organise them into 4 equal groups:
BB = pairs where the siblings are two boys;
BG = older boy and his sister;
GB = older girl and her brother;
GG = two girls.
For the family A scenario, we must have made our random choice from any pair in the first three groups. So it’s clear that 1/3 of our choices will result in two boys.
To make out random choice for family B, we need to ask the siblings where one (or both) boys was born on a Tuesday to stand up and everyone else to sit down. Family B is equally likely to be any of the standing pairs. Obviously, all of GG sit down. In GB and BG, it should be clear that about 1/7 of them will be standing up. But it’s a bit different in BB, where around 2/7 of them will be standing up. (1/7 where older boy born Tuesday, 1/7 where younger boy born Tuesday – it will be slightly less than 2/7 because I’ve double-counted the 1/49 where both were born Tuesday).
This means about half of the pairs who are standing up will be from the BB group. In other words, if we randomly choose from the population of pairs with a boy born on Tuesday, we have around 50% chance of picking from BB, ie both are boys.
Being born on Tuesday is uncommon, so families with two boys have twice the chance of getting a boy born on Tuesday, meaning they get double represention in the families like Family B.