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Probability questions


Wabbit
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I'm curious how many people here can figure these out (don't use the internet! honor system.)

 

1. I have two kids. One is a boy. What is the probability I have two boys?

 

2. I have two kids. One is a boy born on a Wednesday. What is the probability I have two boys?

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Both would be 25% chance. The first condition of having a boy/girl doesn't really matter as the chance of having two of either starts at 25% (1/2 chance compounded once). Propagation of errors in these two 1/2 chances might yield a slightly different answer, but I'm pretty sure its 25%. Unless I'm missing something, the being born on a Wednesday thing shouldn't alter the answer unless you are going by statistical data of what happens when a child is born on a Wednesday. That data would theoretically pan out to having no correlation anyway.

 

Thanks for this, as a Physics major, I need something at least slightly educational during my winter break lol...

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Both would be 25% chance. The first condition of having a boy/girl doesn't really matter as the chance of having two of either starts at 25% (1/2 chance compounded once). Propagation of errors in these two 1/2 chances might yield a slightly different answer, but I'm pretty sure its 25%. Unless I'm missing something, the being born on a Wednesday thing shouldn't alter the answer unless you are going by statistical data of what happens when a child is born on a Wednesday. That data would theoretically pan out to having no correlation anyway.

 

Thanks for this, as a Physics major, I need something at least slightly educational during my winter break lol...

 

Girl Girl

Girl Boy

Boy Girl

Boy Boy

 

since I've told you that one of the children is a boy, you can ignore Girl Girl

 

of the three remaining possibilities, only one (Boy Boy) has the second child as a boy, so the answer is 1/3

the first question is not a difficult one to answer, the actual challenge is in the second one, and it's completely counterintuitive. i'll write up the solution at some point soon.

As for the second question (of course assuming that the births are independent events and that there is an equal probability of a boy and a girl, and an equal probability of being born on each of the seven days of the week):

 

First, we need to map out the complete set of events: but not yet =p

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50%.

 

the reason this isn't just a straight up conditional probability question, is because I did not tell you which child is the confirmed boy. if I had, then obviously the choices for the other child are 50/50.

an easy way to conceptualize it is:

 

imagine we have a (random) group of 1 million families with 2 kids each. if we asked every family with at least one boy to step forward, you'd guess that about 750,000 of them would. then of those families, if we asked those with two boys to step forward, about 250,000 or so of them would.

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As for the second question (of course assuming that the births are independent events and that there is an equal probability of a boy and a girl, and an equal probability of being born on each of the seven days of the week):

 

First, map out the complete set of events:

 

Child A / Child B:

 

Boy on Wednesday / Girl (any of the days) - 7 outcomes

Girl (any of the days) / Boy on Wednesday - 7 outcomes

Boy (any of the days) / Boy on Wednesday - 7 outcomes

Boy on Wednesday / Boy (any of the days) - 6 outcomes (remember: you have counted one of these outcomes in the above scenario)

 

Of these outcomes, 13 fit the proposed scenario - a boy born on Wednesday and another boy. Therefore, the correct answer is 13/27

 

If you need help visualizing this, you can make (or imagine) a 14 x 14 grid of all possible combinations (first child - girl monday, second child - girl monday; first child - girl monday, second child - girl tuesday, etc.) If you then restrict the set to any case that includes a boy born on a Wednesday, you should get 27 cases. Of these, the cases that include another boy should total 13 cases.

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