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Let's see if I can help you understand this a bit better. First, let's clarify the problem being asked. There are 2 different problems with different solutions and it helps to explicitly separate them. problem 1) You go up to a person and ask them if they have exactly 2 children, at least one of which is a boy born on Tuesday. They say yes. What is the probability that they have a girl? problem 2) You go up to a person and ask them if they have exactly 2 children, at least one of which is a boy. They say yes. You then ask them which day of the week a boy they have was born on. They say Tuesday. What is the probability that they have a girl? The original problem that was posed is equivalent to problem 1, but not equivalent to problem 2. This could be what is confusing you, because in problem 2 the extra information plays no role in the selection process, while it does play a role in problem 1. In problem 2, the answer is the standard 2/3. Why are the probabilities different between problem 1 and 2? Here's why: Think about the set of people who could answer yes to the question in problem 2. The ratio of these groups is important. A parent with BB (two boys) is equally likely to answer yes to problem 2 (100% likely to be exact) as a parent with BG and GB (also 100% likely to answer yes), which leads to the correct solution of 2/3. However, in problem 1 a parent with BB is NOT EQUALLY LIKELY to answer yes as a parent with BG. This is because we added an extra qualifier (must be born on Tuesday). The parent with BB has two chances to meet this qualifier because they have two boys, so the parent with BB is actually more likely to answer yes to the question than the parent with BG. As the qualifier becomes more and more rare (day of lunar cycle), the probability of the BB parent answer yes P(yes|BB) approaches twice the value of P(yes|BG). So now you're left with some subset of parents with BB, BG, and GB, but in this scenario you've sampled from BB approximately twice as much as you've sampled from each of the BG and GB groups, leaving you with approximately the same number of people from group BB as the combined amount from groups BG and GB. This is why the probability approaches 50% I spend a while writing this, so hopefully it helps! |
Jones has 2 kids. Let A be "he has a girl" and B be "he has a boy born on tuesday." First thing I realized is A and B are NOT independent - this is key. P(A) includes the option of Jones having two girls. But if B is true, then the two girls option isn't on the table anymore, which affects P(A). Realizing this helped me start to better understand what kind of problem we're dealing with.
Second was realizing that P(A&B) is not at all the same thing as P(A|B) - the probability of A given B - when A and B aren't independent. The problem is asking for P(A|B), and by the rule of conditional probability: P(A|B) = P(A&B)/P(B)
P(B) can be solved for without too much fuss: solve 1-P(!B). For each kid you have 2 genders and 7 days of the week, or 2 * 7 = 14 options. 13 of those are not "Boy & Tuesday." So you have P(!B) is (13/14) * (13/14) = 169/196. P(B) = 1 - 169/196 = 27/196.
This leaves us trying to figure out P(A&B). I can't think of any other way to do it other than enumerating all options. We can take a shortcut and just look at all 27 possible scenarios where B is true. This seems to be the method of choice ;) As others have shown, we see that 14 of those satisfy A. So P(A&B) = 14/196.
Now, we can solve: P(A|B) = P(A&B)/P(B) = (14/196)/(27/196) = 14/27
So I'm now part of the "math checks out" club. Thanks for all the help people!