The answer should be 13/27 with the assumption of having child on Tuesday is equally likely to have child on any other day.
How?
CASE 1. Suppose that the given son born on Tuesday is older guy. So the younger child can either be a boy or a girl. If younger child is boy, there are 7 ways to BE BORN!(on any of the 7 weekdays!) and similarly if the younger child is girl, there are 7 ways to BE BORN!
CASE 2. Now suppose that the older child is not born on Tuesday. So the younger child then must be born on Tuesday, of course. If the older child is girl, then there are 7 ways to BE BORN for her! But now caution: if the older child is a boy, he has only 6 ways to BE BORN! Why? because we have assumed in this case that older child is not born on Tuesday!(And, this case of older+younger guys born on Tuesday was taken up in Case 1 already!)
Now just sum up!
(Ways for guy to be born)/(Total possible ways for the second child to be born)=(7+6)/(7+7+7+6)=13/27
The question depends on how you formulate it. The puzzle is not “My firstborn is a boy born on Tuesday” or “I have exactly one boy born on a Tuesday”, but “One [or more] is a boy born on a Tuesday
On Mon, Mar 17, 2014 at 11:05 AM, SINGHAM [via Discussion forum] <[hidden email]> wrote:
The answer would be 1/3 if the question was:"If I have two children, what is the probability of both of them are boys, given one is a boy?"
Introducing Tuesday has changed the question.
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