The correct answer is not 1/2 but 1/3. How can that be? Well there are four possible combinations, BB, GG, BG and GB but but at least one is a boy so you can get rid of GG. So all that's left is BB, BG and GB; and in only one of those three possibilities do I have two boys.
Now I tell you that I have two children and one of them is a boy born on a Tuesday. What is the probability that I have two boys?
7 comments:
Is the answer 13/27?
I don't know ... why would it be 13/27?
This may be a naive question, isn't day of week the child is born on a non sequitor? I think answer should still be 1/3.
I searched an article containing discussion of this problem and found 13's as numerators but no 27's as denominators.
The answer is not supposed to be 1/3, but it may depend on how you interpret the question.
For example, the answer to the first question can be 1/2 instead of 1/3 with a particular understanding of the statement. If the person "has" two children because they picked them at random from a very large supply of children, half boys and girls, and they have only looked at one child to determine that he is a boy, there is a 1/2 chance that the second child is also a boy.
In other words, with a common-sense understanding of the second statement, it is significant that one of the children was born on a Tuesday.
Your visit today reminded me of this question.
So you have {B,G} for boy/girl and {1,7} for each day of the week, B1 means a boy born on Monday.
All possible combinations for two children are:
{B1,B1}, ..., {B1, B7}, {B2,B1}, ..., {B2,B7}, ...., {B7,B1}, ..., {B7,B7}, {G1,G1}, ..., {G7,G7}
If we eliminate the combinations not including B2 (son born on Tuesday) we get:
{B2,B1}, ..., {B2,B7}, {B2,G1}, ..., {B2,G7}, {B1,B2}, ..., {B7,B2}, ..., {G1,B2}, ..., {G7,B2}
So there are 27 outcomes in total. (not counting {B2,B2} twice). 13 of them is in form {B#,B#}, meaning having 2 boys. Hence the probability 13/27.
Sorry, I was too lazy to figure out your notation and came to the same conclusion through tedious enumeration.
We start with 196 possible outcomes, as you described, starting with 49 ways to have two boys --
B1B1 B1B2 B1B3 B1B4 B1B5 B1B6 B1B7
B2B1 B2B2 B2B3 B2B4 B2B5 B2B6 B2B7
B3B1 B3B2 B3B3 B3B4 B3B5 B3B6 B3B7
B4B1 B4B2 B4B3 B4B4 B4B5 B4B6 B4B7
B5B1 B5B2 B5B3 B5B4 B5B5 B5B6 B5B7
B6B1 B6B2 B6B3 B6B4 B6B5 B6B6 B6B7
B7B1 B7B2 B7B3 B7B4 B7B5 B7B6 B7B7
-- then 49 more for a boy and a girl, 49 more for a girl and a boy, and 49 more for two girls.
In 27 of these cases, we see a boy born on a Tuesday:
B1B2
B2B1 B2B2 B2B3 B2B4 B2B5 B2B6 B2B7
B3B2
B4B2
B5B2
B6B2
B7B2
B2G1 B2G2 B2G3 B2G4 B2G5 B2G6 B2G7
G1B2
G2B2
G3B2
G4B2
G5B2
G6B2
G7B2
In 13 of these 27 cases, we observe two boys.
I did not find 13/27 mentioned as the answer in the article I searched because it is expressed as the probability of the other child being a girl, and in decimal notation.
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