There’s a post today over at coding horror that proposed a very simple probability problem. After more than 400 comments (and counting) I’m very surprised that so many people have the wrong answer. The question is this:
Let’s say, hypothetically speaking, you met someone who told you they had two children, and one of them is a girl. What are the odds that person has a boy and a girl?
Actually, I’m not sure if by “odds” Jeff meant this, or probability — as is the case in most everyday use. But assuming that he meant probability, then the solution is as follows.
Let A be the event of having a girl.
Let B be the event of having a boy.
Assume that P(A) = P(B) = 0.5
We know that the person has a girl, thus we are trying to find P(B|A) (probability of B given that A has occurred).
Now we note that A and B are independent events, thus P(B|A) = P(B) = 0.5 (conditional probability).
And there you go, the probability of the person having a boy given that person has a girl is 0.5 (or 50%).
Happy New Year’s folks!
April 10th, 2011 at 10:26 pm
Your answer is wrong. You are calculating the probability that another child will be a boy, which is of course 0.50, but that is not what was asked. The person you met had two children. There are four possbilities for this: 1st boy 2nd boy; 1st boy 2nd girl; 1st girl 2nd boy; and 1st girl 2nd girl. You see that, if a person has two children, the chance is 0.75 that at least one is a boy. (You can also figure this out by calculating the odds of two girls, 0.5 x 0.5 = 0.25, the only situation in which there is no boy.) Now, since you have been told that one child is a girl, the boy-boy possibility is eliminated, and you are left with three: boy-girl, girl-boy, girl-girl. This means that the odds that the person has a boy and a girl is 0.67, not 0.50. This is assuming, of course, that the person has told you the truth.