This is a cute little problem - one that looks straightforward when it isn't!
boy-or-girl.txt
Mary, Bob, and Alice are playing a noisy game in the park where you are trying to read. The two girls are as boisterous as the boy ruining your hopes of a quiet afternoon. You look up with a sigh and the woman sitting next to you catches your eye.
"I do hope my children aren't disturbing you." she says.
"It's alright," you reply politely, "It must be hard to manage three kids at that age."
"Oh only two are mine!" she laughs, "Alice and ______"
What is the probability that
* she ends with "Bob" = ___%
* she ends with "Mary" = ___%
Think about it!
Try and be sure of the answer before checking out the solution!
UPDATE: 2009-01-15
Amit Bali pointed out that it had to be made clear the woman had no kids at home. So in case you were wondering if the woman had more kids stashed away at home - she doesn't!
Thanks Amit!
!!!!!!!!!!!!!!!!!!!! Solution !!!!!!!!!!!!!!!!!!!!
Initially it may seem obvious that the answer is 50%. After all, the woman could have either a boy or a girl so that's an even chance. But let's look at the problem a little deeper.
The woman has told us that she has a little girl. How can we use this information?
We can work out the solution mathematically. It is quite simple but not obvious. We'll do that so we are confident of our answer:
The set of possible children the woman has is:
{BB, BG, GB, GG}
Out of this we select those that filter out no girls leaving:
{BG, GB, GG}
Hence the chances of having a boy (Bob) is:
|{BG,GB}|/|{BG,GB,GG}| = 2/3 or ~67%
while the chances of having a girl (Mary) is:
|{GG}|/|{BG,GB,GG}| = 1/3 or ~33%
So the probability that
* she ends with "Bob" = ~67%
* she ends with "Mary" = ~33%
The math is clear then but this seems very unintuitive! In fact, it is so counter-intuitive that people figure that the
{BG,GB} permutation is incorrect and should "collapse" (into just
{BG} or
{GB}) thus giving us the 50% we are expecting.
So let's try another explanation of the solution.
First try to remember that you do not know this woman. I.e. she is a random sample from the entire population of your city. Keep this in mind and the explanation gets easier.
You with me? This is a random woman so there is a probability you can apply to her as well. Let's see what this is.
Say there are 100 families lined up, all of whom have two children. Now there are 50% boys and 50% girls in the group.
All hunky-dory so far?
There are 50% boys and 50% girls in the group
BUT there also are families which have
no girls (two-boy families) and families that have
no boys (two girl families).
Got that? 100 families, some with only boys, some with only girls, the rest with both boys and girls.
Now it should be obvious that the no-girls families and the no-boys families are less than the boy-girl families. This is because the boy-girl families can have an older girl and a younger boy
OR a younger girl and an older boy!
Since the chances of having a girl or a boy are even, the families nicely distribute into:
- 25% only boy families
- 25% only girl families
- 25% girl/boy families
- 25% boy/girl families
Now, our woman is some random stranger from this group! Let's keep that in mind and ask families in the group with at least one girl to step forward.
3 groups step forward (all except the only boy families).
Our woman is therefore from one of these 3 groups. She cannot be from the all boys group.
Within these three groups, two have a second boy (2/3) and only one has a girl (1/3). Thus we can say with 2/3 or ~67% probability that the woman has a boy and a 1/3 or ~33% probability that she has another girl.
Phew! I hope you followed all that!
!!!!!!!!!!!!!!!!! Solution Ends !!!!!!!!!!!!!!!!!!!!
