# I love probability, I love probability, I love probability, ... (Assume child has...)

#### Jomo

##### Elite Member
Assume that a child has an equal chance of being born on any month and of either gender (male, female). Let p be the probability that given that a family has two children and at least one of them is a girl, both are girls. Furthermore, let q be the probability that given that a family has two children and at least one of them is a girl born in April, that both are girls. Does p=q? The answer is no but I do not see why yet.

Here is what I have done so far.

The 1st scenario. The family has two children and at least one is a girl. So the family has either gb, bg or gg. In one of the three cases there is a 2nd girl. So p=1/3. I am certain that this is correct.

2nd case. The family has two children and at least one is a girl born in April.

The family can have either bg(girl in April), gb(girl in April), gg (1st girl in April) or gg (2nd girl in April). Note that the 'other girl' (if there is one) could have been born in April as well.

Now in the 1st case the given girl could have been born in any of the 12 month and in this case we are told April.

So is the answer for q, q= 2/4 = 1/2 because we have 4 cases in the 2nd scenario? Well that is only true if the four possibilities are all equally likely. I am not totally convinced that that is true.

Any takers on how to explain this? Thanks

#### Dr.Peterson

##### Elite Member
Assume that a child has an equal chance of being born on any month and of either gender (male, female). Let p be the probability that given that a family has two children and at least one of them is a girl, both are girls. Furthermore, let q be the probability that given that a family has two children and at least one of them is a girl born in April, that both are girls. Does p=q? The answer is no but I do not see why yet.

Here is what I have done so far.

The 1st scenario. The family has two children and at least one is a girl. So the family has either gb, bg or gg. In one of the three cases there is a 2nd girl. So p=1/3. I am certain that this is correct.

2nd case. The family has two children and at least one is a girl born in April.

The family can have either bg(girl in April), gb(girl in April), gg (1st girl in April) or gg (2nd girl in April). Note that the 'other girl' (if there is one) could have been born in April as well.

Now in the 1st case the given girl could have been born in any of the 12 month and in this case we are told April.

So is the answer for q, q= 2/4 = 1/2 because we have 4 cases in the 2nd scenario? Well that is only true if the four possibilities are all equally likely. I am not totally convinced that that is true.

Any takers on how to explain this? Thanks
You haven't worked it out at all. Take into account that the probability of being a girl born in April is 1/24, and so on, and you will get a probability that is not 1/3; in fact, it is close to 1/2. Nothing can be assumed to be equally likely.

I did it by making a table with three possibilities for each child (boy, non-April girl, April girl) and filling in all the 9 probabilities, then doing the conditional.

#### Jomo

##### Elite Member
You haven't worked it out at all. Take into account that the probability of being a girl born in April is 1/24, and so on, and you will get a probability that is not 1/3; in fact, it is close to 1/2. Nothing can be assumed to be equally likely.

I did it by making a table with three possibilities for each child (boy, non-April girl, April girl) and filling in all the 9 probabilities, then doing the conditional.
Ok, so I look at it again right here.
The family can have either bG, Gb, Gg or gG where G is a girl born in april.
So $$\displaystyle P(bG) = \frac{1}{2}\frac{1}{24} = \frac{1}{48}, P(Gb) = \frac{1}{24}\frac{1}{2} = \frac{1}{48}, P(Gg) = \frac{1}{24}\frac{1}{2} = \frac{1}{48}, P(gG) = \frac{1}{2}\frac{1}{24} = \frac{1}{48}$$. Now from what I see they are all equal (to 1/48).

I am still not there yet.

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#### Dr.Peterson

##### Elite Member
Ok, so I look at it again right here.
The family can have either bG, Gb, Gg or gG where G is a girl born in april.
So $$\displaystyle P(bG) = \frac{1}{2}\frac{1}{24} = \frac{1}{48}, P(Gb) = \frac{1}{24}\frac{1}{2} = \frac{1}{48}, P(Gg) = \frac{1}{24}\frac{1}{2} = \frac{1}{48}, P(gG) = \frac{1}{2}\frac{1}{24} = \frac{1}{48}$$. Now from what I see they are all equal (to 1/48).

I am still not there yet.
What is P(g)? It isn't 1/2 as you are thinking; "g" is a girl who was NOT born in April, right? Half of all children are not girls not born in April ...

#### Jomo

##### Elite Member
What is P(g)? It isn't 1/2 as you are thinking; "g" is a girl who was NOT born in April, right? Half of all children are not girls not born in April ...
I am sure that you are right but my thinking is that the other girl could also have been in April. All you are given is that one of two children is a girl who was born in April. So I do think that p(g) = 1/2. Why should the other girl not be born in April?

Half of all children are not girls not born in April: There are 12 month per year for each male (or female) to be born in. So p(boy not being a girl not born in April) = 1. P(girl not being a girl born in April) = 11/12. Now p(a child not being a girl born not born in April) = .5( 1 + 11/12) = 23/24 = 1 - p(a child being a girl who was born in April). So this seems right as I got 23/24 doing it two ways.

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#### Dr.Peterson

##### Elite Member
I am sure that you are right but my thinking is that the other girl could also have been in April. All you are given is that one of two children is a girl who was born in April. So I do think that p(g) = 1/2. Why should the other girl not be born in April?

Half of all children are not girls not born in April: There are 12 month per year for each male (or female) to be born in. So p(boy not being a girl not born in April) = 1. P(girl not being a girl born in April) = 11/12. Now p(a child not being a girl born not born in April) = .5( 1 + 11/12) = 23/24 = 1 - p(a child being a girl who was born in April). So this seems right as I got 23/24 doing it two ways.
I get the impression that you are not being careful about definitions. "g" is not "whatever the other girl is"; it's "a girl not born in April", at least as I'm taking it. The probability of that can't be the same as "a girl"; it has to be less, specifically, P(g) = 1/2 * 11/12 = 11/24. You have calculated P(not being a girl born in April), which is 23/24. Neither of these is 1/2.

I made a grid listing all 9 possibilities, where b=boy, g=girl not born in April, G=girl born in April. These are bb, bg, bG, gb, gg, gG, Gb, Gg, GG. Then we are given that it is either bG, gG, Gb, Gg, or GG -- you missed GG last time! And we want the conditional probability that it is gG, Gg, or GG.

#### Jomo

##### Elite Member
I get the impression that you are not being careful about definitions. "g" is not "whatever the other girl is"; it's "a girl not born in April", at least as I'm taking it. The probability of that can't be the same as "a girl"; it has to be less, specifically, P(g) = 1/2 * 11/12 = 11/24. You have calculated P(not being a girl born in April), which is 23/24. Neither of these is 1/2.

I made a grid listing all 9 possibilities, where b=boy, g=girl not born in April, G=girl born in April. These are bb, bg, bG, gb, gg, gG, Gb, Gg, GG. Then we are given that it is either bG, gG, Gb, Gg, or GG -- you missed GG last time! And we want the conditional probability that it is gG, Gg, or GG.
I am confused about what you are saying. You are saying that g is a girl not being born in April. Yet you speak of (my not) looking at GG. Clearly GG must mean that both girl are born in April. So why not consider Gg and gG where g can be born in April (but does not have to be)?

I understand that two people can solve a probability problem 'correctly' even though the results are different. This is because the people solving the problem do not agree with the interpretation of the problem. Now I am not saying that the way that I am doing it is correct via my interpretation. Now I respect you greatly as a mathematician and enjoy learning from you so I am asking why you think that the other girl can't be born in April. And then you talk about GG which does imply that the other girl (in this case) is born in April.

I apologize for not understanding this more quickly.

#### Dr.Peterson

##### Elite Member
I am confused about what you are saying. You are saying that g is a girl not being born in April. Yet you speak of (my not) looking at GG. Clearly GG must mean that both girl are born in April. So why not consider Gg and gG where g can be born in April (but does not have to be)?

I understand that two people can solve a probability problem 'correctly' even though the results are different. This is because the people solving the problem do not agree with the interpretation of the problem. Now I am not saying that the way that I am doing it is correct via my interpretation. Now I respect you greatly as a mathematician and enjoy learning from you so I am asking why you think that the other girl can't be born in April. And then you talk about GG which does imply that the other girl (in this case) is born in April.

I apologize for not understanding this more quickly.
The issue here is not one of interpretation, but of definition.

We have to define what we mean by "g". You can't mean two things by it, or you'll confuse yourself.

You appear to be defining it as "a girl who is not necessarily born in April". But if you do that, then g and G are not mutually exclusive, and you can't add probabilities. In particular, your Gg and gG would both include GG, and the probability of "Gg or gG" would have to be P(Gg) + P(gG) - P(GG). That would make things much too complicated. Take g to mean "a girl who is born in a month other than April", and everything becomes clear.

It is essential (at least the way we are approaching the problem) to partition the sample space, so that we have a set of events that do not overlap, and that together cover all possibilities. That is what I have done by defining the events b, g, and G as I have.

... so I am asking why you think that the other girl can't be born in April. And then you talk about GG which does imply that the other girl (in this case) is born in April.
I don't think "the other girl can't be born in April", and I never said anything of the sort. In including GG as a possibility, I am explicitly allowing that! You, on the other hand, by ignoring that, are not -- or rather, you are confusing the issue by not being consistent in what you mean by "g". What did I say that you are interpreting as saying that only one can be born in April???

#### Jomo

##### Elite Member
The issue here is not one of interpretation, but of definition.

We have to define what we mean by "g". You can't mean two things by it, or you'll confuse yourself.

You appear to be defining it as "a girl who is not necessarily born in April". But if you do that, then g and G are not mutually exclusive, and you can't add probabilities. In particular, your Gg and gG would both include GG, and the probability of "Gg or gG" would have to be P(Gg) + P(gG) - P(GG). That would make things much too complicated. Take g to mean "a girl who is born in a month other than April", and everything becomes clear.

It is essential (at least the way we are approaching the problem) to partition the sample space, so that we have a set of events that do not overlap, and that together cover all possibilities. That is what I have done by defining the events b, g, and G as I have.

I don't think "the other girl can't be born in April", and I never said anything of the sort. In including GG as a possibility, I am explicitly allowing that! You, on the other hand, by ignoring that, are not -- or rather, you are confusing the issue by not being consistent in what you mean by "g". What did I say that you are interpreting as saying that only one can be born in April???
With some help and some deep thinking I am very confidence that I now understand this problem.

Here is my solution.

Let the event A = family has one girl born in April
event B =family has two girls

There are 12x12 = 144 possible combinations for the months that the two girls can be born on.

Exactly 12x12 - 1 = 23 of these 144 have a girl born in April. So P(A|B) = 23/144

We know that P(B) = 1/4

Now P(B|A) = P(A|B)P(B)/P(A). We only need P(A) to finish up.

There are (2x12)2 = 576 ways to get the month and gender of the two children.

Note that 232 = 529 ways do not have a girl born in April, so 576 - 529 = 47 do.

Then P(A) = 47/576

Now P(B|A) = P(A|B)P(B)/P(A) = (23/144)(1/4)/(47/576) = 23/47

I do not have the answer key for this problem. Is this the answer you got? Also do you see any problem with my solution?

I do see that gG and Gg included GG twice, since you pointed it out.

#### Dr.Peterson

##### Elite Member
With some help and some deep thinking I am very confidence that I now understand this problem.

Here is my solution.

Let the event A = family has one girl born in April
event B =family has two girls

There are 12x12 = 144 possible combinations for the months that the two girls can be born on.

Exactly 12x12 - 1 = 23 of these 144 have a girl born in April. So P(A|B) = 23/144

We know that P(B) = 1/4

Now P(B|A) = P(A|B)P(B)/P(A). We only need P(A) to finish up.

There are (2x12)2 = 576 ways to get the month and gender of the two children.

Note that 232 = 529 ways do not have a girl born in April, so 576 - 529 = 47 do.

Then P(A) = 47/576

Now P(B|A) = P(A|B)P(B)/P(A) = (23/144)(1/4)/(47/576) = 23/47

I do not have the answer key for this problem. Is this the answer you got? Also do you see any problem with my solution?

I do see that gG and Gg included GG twice, since you pointed it out.
Here is my solution, using b=boy, g=girl not in April, G=girl in April:

 b g G total b 1/4 11/48 1/48 1/2 g 11/48 121/576 11/576 11/24 G 1/48 11/576 1/576 1/24 total 1/2 11/24 1/24 1

Event A (at least one girl born in April) is underlined, and event B (two girls) is in bold.

So P(B | A) = P(A and B)/P(A) = (11/576 + 11/576 + 1/576)/(1/48 + 1/48 + 11/576 + 11/576 + 1/576) = (23/576)/(47/576) = 23/47. So you are right.

As I said, this is close to 1/2.

Your solution looks fine.

#### Jomo

##### Elite Member
Here is my solution, using b=boy, g=girl not in April, G=girl in April:

 b g G total b 1/4 11/48 1/48 1/2 g 11/48 121/576 11/576 11/24 G 1/48 11/576 1/576 1/24 total 1/2 11/24 1/24 1

Event A (at least one girl born in April) is underlined, and event B (two girls) is in bold.

So P(B | A) = P(A and B)/P(A) = (11/576 + 11/576 + 1/576)/(1/48 + 1/48 + 11/576 + 11/576 + 1/576) = (23/576)/(47/576) = 23/47. So you are right.

As I said, this is close to 1/2.

Your solution looks fine.
I see clearly that you used g as girl not born in April so that you can use GG to show that both girl could be born in April. Your table is very nice. Thanks!