Not really - like Norvig said, that's one interpretation, but there is another valid one. Say you had a bunch of families with two children each. The children

Not really - like Norvig said, that's one interpretation, but there is another valid one. Say you had a bunch of families with two children each. The children are evenly distributed in terms of gender and the the days of the week on which they were born. If you pick one parent from the crowd, the chance that they have at least one boy is 3 / 4, the chance that they have two is 1 / 4, and the chance that they have none is also 1 / 4: | B| G| --|--|--| B |BB|BG| --|--|--| G |GB|GG| Each of the four squares on the above table is equiprobable. However, if the person says they have at least one boy, they must be in either the left column or the top row, so in one of three squares. There is only one square in those three with both boys, so the chance of that parent having two boys is 1 / 3. Now, for the day-of-the-week problem. If you ask a parent if they have a male child born on Tuesday, it is not equiprobable that they're in any of those three squares. In the BG group, they all have male children, so the chance that any parent chosen has a male child born on Tuesday is 1 / 7. Similarly in the GB group. However, in the BB group, either one of their children may be born on a Tuesday to satisfy the condition. The chance that either child is born on a Tuesday is the same as the inverse of neither child being born on a Tuesday, or 1 - (6/7 * 6/7) = 13/49 So the number of parents in the top left group (BB) who satisfy the condition is 13 / 49, whereas the number of parents in the top right (BG) is 1 / 7, and the bottom left (GB) is also 1 / 7. You're looking for the probability that a given parent in that satisfies the condition is in the top left group, which is (13/49) / (1/7 + 1/7 + 13/49) = 13/27