today's fun with probability
Nov. 20th, 2002 07:33 pm![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
eub1) You meet a man at the bowling alley [this is how it was posed, honest]. He has two children, bowling on lane 17. You see one girl there; the other child doesn't happen to be visible.
2) Man, bowling, two children. You ask if he has any daughters. He says yes.
In each case (any difference?), what's the probability that he has two daughters?
Assume male and female are mutually exclusive and 50-50 and independent between children. Make any further assumptions you need.
[edit: spoilers in comments, natch]
2) Man, bowling, two children. You ask if he has any daughters. He says yes.
In each case (any difference?), what's the probability that he has two daughters?
Assume male and female are mutually exclusive and 50-50 and independent between children. Make any further assumptions you need.
[edit: spoilers in comments, natch]
They're different!
Date: 2002-11-21 12:19 pm (UTC)In the first situation, there's a 50% chance of having two girls, and in the second, it's 1 in 3.
A good way to think about this is by writing out all of the possibilities and seeing which ones you can rule out. There are 4 equally likely child-pair possibilities:
A older and younger are girls
B older is boy and younger is girl
C older is girl and younger is boy
D older and younger are boys
In case 1, there are 4 things which could have occurred:
either we have pair A and we saw the older girl
we have pair A and we saw the younger girl
we have pair B and we saw the younger (only) girl
we have pair C and we saw the older (only) girl
Since each of these situations is equally likely, there's a 2 in 4 chance (50%) that both are girls.
In case 2, we can rule out possibility D (2 boys), which leaves us with A, B, and C to choose from. Out of those 3 possibilities, only one has two girls, so the chance that there are two girls is 1 in 3.
The reason that these two situation are different is that in the first, we can fix the gender of a child that is chosen at random from the two, whereas the second situation does nothing to specify the child.
If you'd like to experiment with the second situation, try flipping two coins, and count how many times there is at least one head as well as how many times there are two heads. If you divide the # of 2 heads by # of at least one head, you should get a ratio close to 1/3.
--A Random Probabilist
Re: They're different!
Date: 2002-12-06 04:07 am (UTC)Choose(2,2)/{ 2^2-Choose(2,0) } == 1/3