Yeah, it is.
Good problem for an actuary...too early in the morning for me.
Your probability is 100%, or 1.
It’s an easy problem. Write out the possibilities.
Good article; thanks for posting!
It’s all about how you ask the question.
If your son was born last Tuesday, the probability you have two sons is 2/3.
Complete baloney! The author correctly states that, for probability calculations, “The first—and most important—rule of counting is this: What is everything that can happen? In the “Mrs Smith” problem, given the information provided, everything that can happen is this:
Boy, Girl
Girl, Boy
Boy, Boy.”
The list should include all the DIFFERENT possibilities. Possibilities 1 and 2 are the same for purposes of the problem. Therefore the probability is (close to) 1/2, not 1/3.
It doesn’t say “only” one born on Tuesday.
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You are very blessed, at least until they reach their teens and college ages.
If you had one child on a tuesday, and it was a boy, the odds that a second child born would be a boy would be one in two. The probability of a future event is not affected by a past event.
But that’s not what this topic is about. What we’re asking is “of those families with two children and at least one boy, how many have two boys?” There are two possibilities for one girl and one boy (first child is a girl, or the second child is a girl), and only one possibility for two boys.
I would disagree. Odds are 1/3 no matter if the day of the week is specified.
This is a variation of the Monty Hall Problem:
http://en.wikipedia.org/wiki/Monty_Hall_problem
This is less a "tricky" probability question than a tricky and ambiguous grammar question. So I cry foul.
Whatever the other child is, it is NOT a son born on Tuesday.
So what are the possibilities?
It could be a son born on one of the other six days of the week.
Or it could be a daughter born on any of the seven days of the week.
So that’s 13 possible outcomes. Six of those outcomes leads to two sons.
Whatever the percentage is, the answer is 6/13
13/27 if you mean at least one boy born on a Tuesday. 6/13 if you mean one and only one boy born on a Tuesday.
Just figure the odds of each of the three sex/day combinations:
Boy Tuesday (bt) = 1/14
Boy not Tuesday (bnt) = 6/14
Girl (g) = 7/14 = 1/2
Form a table of all nine ordered combinations (bt*bt = 1/196, bt*bnt = 6/196, bnt*bt = 6/196, etc.) and throw out the ones which don't have at least one bt (maybe also throw out the one with two bt depending on whether you want one and only one boy born on Tuesday). Then take the sum of the pairs with two boys (13/196) divided by the sum of all allowed combinations (27/196) and get the answer (13/27).
Hey! I’m getting ready to go on vacation tomorrow. Now my head hurts...
There is no probability, since both have already been born.
Boy, Girl
Boy, Boy
Girl, Boy
Boy, Boy
The “born on a Tuesday” is irrelevant to the question asked.
One child is a son.
The other child is either a son or a daughter.
Assume for the sake of this discussion that the birth of boys and girls is of equal probability.
The probability that the second child is a son is 1/2.
Variations on the question such as “I have two children, one a son born on Tuesday; what is the probability that my firstborn is a boy?” have different answers.
The answer to the question as posed is “1/2”.