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# One Girl - One Boy

## Question

Ok, so Teanchi and Beanchi are a married couple (dont ask me whose he and whose she)!

They have two kids, one of them is a girl, what is the probability that the other kid is also a girl.

Assume safely that the porbability of each gender is 1/2.

Ofcourse its not 1/2 else would make it a lousy puzzle...

Ans: 1/3

This is a famous question in understanding conditional probability, which simply means that given some information you might be able to get a better estimate.

The following are possible combinations of two children that form a sample space in any earthly family:

Girl - Girl

Girl - Boy

Boy - Girl

Boy - Boy

Since we know one of the children is a girl, we will drop the Boy-Boy possibility from the sample space.

This leaves only three possibilities, one of which is two girls. Hence the probability is 1/3

## 348 answers to this question

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*sigh*

i'm sorry, but this is a sad attempt at statistics. The fact of the matter is, you are trying to figure the probability of the 2 kids before you know the outcome of the first.

Before we know the first kid is a girl, the probability is 25% that both are girls.

Once we know that the first is a girl, you say we have to remove the boy-boy from the possibilites. Thus you say that we now have girl-boy, girl-girl, and boy-girl. But this last one must also be eliminated because we don't have boy as the first child. Thus, only girl-boy and girl-girl are our only two answers. 50%

But that does matter, cause it doesn't effect the outcome of the second AT ALL. And here is my reason: regardless of what was released during the first conception, half of the mans sperm will contain an X chromosome, and the other half a Y. And if it isn't exactly half, then its too close to count.

So believe your faulty statistics, I'm gonna go with biology on this one...

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*sigh*
i'm sorry, but this is a sad attempt at statistics. The fact of the matter is, you are trying to figure the probability of the 2 kids before you know the outcome of the first.

Before we know the first kid is a girl, the probability is 25% that both are girls.

Once we know that the first is a girl, you say we have to remove the boy-boy from the possibilites. Thus you say that we now have girl-boy, girl-girl, and boy-girl. But this last one must also be eliminated because we don't have boy as the first child. Thus, only girl-boy and girl-girl are our only two answers. 50%

But that does matter, cause it doesn't effect the outcome of the second AT ALL. And here is my reason: regardless of what was released during the first conception, half of the mans sperm will contain an X chromosome, and the other half a Y. And if it isn't exactly half, then its too close to count.

So believe your faulty statistics, I'm gonna go with biology on this one...

Unfortunately, it is not.

The *sigh* and the *tear* apply -- there is sadness here -- but not in the way you might be thinking.

Consider the plight of the OP -- the one who wrote the riddle.

The riddle that so many have simply not taken the time to read.

Read the OP once again - this time carefully.

You will see that we do not know the gender of the "first" child.

Something so trivial as "what's the likelihood of my second child being a girl?" HARDLY constitutes a riddle, does it?

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Ok, so Teanchi and Beanchi are a married couple (dont ask me whose he and whose she)!

They have two kids, one of them is a girl, what is the probability that the other kid is also a girl.

Assume safely that the porbability of each gender is 1/2.

Ofcourse its not 1/2 else would make it a lousy puzzle...

Ans: 1/3

This is a famous question in understanding conditional probability, which simply means that given some information you might be able to get a better estimate.

The following are possible combinations of two children that form a sample space in any earthly family:

Girl - Girl

Girl - Boy

Boy - Girl

Boy - Boy

Since we know one of the children is a girl, we will drop the Boy-Boy possibility from the sample space.

This leaves only three possibilities, one of which is two girls. Hence the probability is 1/3

Hello guys

In my opinion the probability is 1/2 because in this case the order doesn't count...It's like the difference between combinations and arrangements:in combinations the order is not important while in arrangements it obviously matters.As i understand this problem, there are only 2 possible cases:girl-girl and girl-boy (or boy-girl,as i said the order doesn't matter).

Edited by bonanova

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Hello guys
In my opinion the probability is 1/2 because in this case the order doesn't count...It's like the difference between combinations and arrangements:in combinations the order is not important while in arrangements it obviously matters.As i understand this problem, there are only 2 possible cases:girl-girl and girl-boy (or boy-girl,as i said the order doesn't matter).

Hi Arkantos,

Your answer requires that the likelihood of a two-girl family is the same as the likelihood of a one-boy and one-girl family.

Do you really believe those are equally likely events?

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Hi Arkantos,

Your answer requires that the likelihood of a two-girl family is the same as the likelihood of a one-boy and one-girl family.

Do you really believe those are equally likely events?

Uhmm,no...i'm just saying that the cases boy-girl and girl-boy in fact represent only one case because the puzzle was clearly saying "They have two kids, one of them is a girl, what is the probability that the other kid is also a girl.".When i was talking about the order i was referring only to these two cases (girl-boy, boy-girl), i wasn't talking about all 3 cases.Sorry if it appeared confusing...my bad

Edited by Arkantos

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Uhmm,no...i'm just saying that the cases boy-girl and girl-boy in fact represent only one case because the puzzle was clearly saying "They have two kids, one of them is a girl, what is the probability that the other kid is also a girl.".When i was talking about the order i was referring only to these two cases (girl-boy, boy-girl), i wasn't talking about all 3 cases.Sorry if it appeared confusing...my bad

I think it's a pretty slippery puzzle. If you were to single out either of the kids and say "that one's a girl", then the probability of the other being a girl is 1/2. The OP doesn't actually state that this kid or that kid is a girl, only that one of them is. That would be true regardless of whether the girl is the first or second kid. So all 3 cases girl-girl, boy-girl, girl-boy apply. Very slippery though.

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I think it's a pretty slippery puzzle. If you were to single out either of the kids and say "that one's a girl", then the probability of the other being a girl is 1/2. The OP doesn't actually state that this kid or that kid is a girl, only that one of them is. That would be true regardless of whether the girl is the first or second kid. So all 3 cases girl-girl, boy-girl, girl-boy apply. Very slippery though.

I have understood this solution from the very beginning,but i still think that the order doesn't matter in this problem....it's a matter of interpretation and i won't try to convince anyone with my solution.

Anyways,let me put it this way:you have two cards with their faces down on a table,their faces are either white either black but you know that at least one of them is black.Which is the probability that the other card is black too?

It's the same thing.Now tell me,if one of them is black and the other is white,does it matter if the black card is in the right side or the left side?

Edited by Arkantos

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Anyways,let me put it this way:you have two cards with their faces down on a table,their faces are either white either black but you know that at least one of them is black.Which is the probability that the other card is black too?
As you've worded it, 1/3

It's the same thing.Now tell me,if one of them is black and the other is white,does it matter if the black card is in the right side or the left side?
That depends:

Me: Is there a black card?

You: Yes

Me: Where?

You: Right hand side

The probability of the card on the left being black is 1/3

Me: What colour is the card on the right hand side?

You: Black

The probability of the card on the left being black is 1/2

Me: Tell me something

You: The card on the right is black

The probability of the card on the left being black depends on the details. If you looked at only the card on the right before saying that, I'd say 1/2. If you looked at both, it may be 1/3 depending on what rules determine the relationship between what the cards are and what you tell me.

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As you've worded it, 1/3

That depends:

Me: Is there a black card?

You: Yes

Me: Where?

You: Right hand side

The probability of the card on the left being black is 1/3

Me: What colour is the card on the right hand side?

You: Black

The probability of the card on the left being black is 1/2

Me: Tell me something

You: The card on the right is black.The probability of the card on the left being black depends on the details. If you looked at only the card on the right before saying that, I'd say 1/2. If you looked at both, it may be 1/3 depending on what rules determine the relationship between what the cards are and what you tell mae.

Nice...

"Now tell me,if one of them is black and the other is white,does it matter if the black card is in the right side or the left side?"-I wasn't asking you which was the probability,it was a "yes or no" question because i wanted to show you that the arrangement in the case when one's black and the other is white wouldn't count when you calculate the probability. From your post i deduce that you would have answered yes,the order does count.Well...i give up . I can't explain this,at least not in english and not on a forum.

PS:And you weren't supposed to solve it by asking me questions,you're in the second major case(the first one being black-black), knowing that there's a black card and a white card (or a boy and a girl) and if the arrangement does not count as i'm saying then we have two cases: black black or black white (=white black)=> probability 1/2....sorry,i wasn't clear and maybe i have some fixed ideas

PPS:And those situations from your post would all have the same answer if i look at it from your point of view (as a probability problem):1/2,the 1/3 answer is false

Edited by Arkantos

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[ Better. The first guess you have a 1 in 3 chance of being right. Once I remove one, your odds move up to 2 in 3 by selecting the OTHER box, because it would have been like selecting the 2 boxes you didn't pick.

If I had offered for you to trade your ONE pick for the TWO you didn't pick, before showing one was empty would you? ]

Wrong, when you remove one. My chances simply becomes 1 in 2.

It just I was ask to choose from 1 in 3, and when you take 1 away it becomes a choice of 1 in 2.

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Uhmm,no...i'm just saying that the cases boy-girl and girl-boy in fact represent only one case because the puzzle was clearly saying "They have two kids, one of them is a girl, what is the probability that the other kid is also a girl.".When i was talking about the order i was referring only to these two cases (girl-boy, boy-girl), i wasn't talking about all 3 cases.Sorry if it appeared confusing...my bad

It's true that there are two cases: [1] both are girls and [2] there is one of each. One of those cases [the first] is the favorable one. The OP asks, what is the probability of [1]? You answer that it must be 1 out of 2 because 1 case out of 2 cases is favorable. What I want to help you to see is that counting the number of cases gets you into all kinds of trouble unless you do one more thing.

To see what that is, try this thought experiment: You buy a lottery ticket. How many outcomes are there? The answer is two: [1] it's the winning ticket, or [2] it's a losing ticket. What is the probability your ticket is the winning ticket? Count the cases: 2. Count the favorable cases: 1. Therefore your probability of winning the lottery is 1 out of 2. Ooops.

When I asked you if you believed that mixed-gender 2-child families and 2-girl 2-child families had equal likelihood, I wasn't concerned about wording or how you described the boy-girl or girl-boy order. I was trying to help you remind yourself that those probabilities are not equal: p[girl-girl] = 1/2 p[girl-boy, boy-girl]. The lottery ticket experiment is an easy way to see that counting up cases that aren't equally likely leads you into error.

The reason that it's helpful to consider the boy-girl and girl-boy cases separately is that then you can count up all the cases: [1] GG, [2] GB [3] BG. Why? Because these three cases have equal probabilities. And since they are the only possible cases, their probabilities add up to 1.

p[GG] + p[GB] + p[bG] = 1.

Now do the math, and find p[GG].

Does that help?

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The solution is incorrect. The probability of a second child does not depend on the probability of the first child. If you flip a coin and it turn out to be heads, the probability that the next flip will be heads is still 1/2. Check any statistics book.

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The solution is incorrect. The probability of a second child does not depend on the probability of the first child. If you flip a coin and it turn out to be heads, the probability that the next flip will be heads is still 1/2. Check any statistics book.

lsmithers, please go through the trouble of reading enough posts in this thread so you can understand why the solution given isn't incorrect. The riddle does not ask about "the second" child and reasons why the riddle is not analogous to what the probability of a second flip of a coin will be has been given multiple times.

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Dear all,

The solution is definitely incorrect, the answer is 50% without a doubt.

First to prove the solution wrong:

There are only 4 possibilies of the combination, namely:

Boy, Boy 25%

Girl, Girl 25%

Boy, Girl 25%

Girl, Boy 25%

If we were to take these possibilies as a set, the chances of getting a boy with a girl is 50% whereas boy boy and girl girl are 25% each. Noted that this is where the person who comes out with the wrong solution is misleaded.

BUT look carefully, when the first option is already out. Say its a girl in this case.

Boy, Boy 25%

Girl, Girl 25%

Boy, Girl 25%

Girl, Boy 25%

We will be left with only 2 combinations, it becomes:

Girl, Girl 50%

Girl, Boy 50%

Hope you can see the picture now. Regardless whether it is a boy or a girl first, the 2nd option become 50%. In the case of Boy first, it becomes:

Boy, Boy 50%

Boy, Girl 50%

Its rather ironic that I have to go into such details into explaining why the solution is wrong, because the logic is just so simple:

Now to show why 50% is the correct answer:

If i give you 10 boxes to choose to from, and only 1 has a treasure in it. You have 10% chance of getting the box wth treaure. But if i take out 6 and leave you 4 boxes to choose from, with a treasure in 1 of the 4 boxes left, you will have 25% chance.

The point is simple, calculate with what you have. In this case, the first girl is out of the picture. We are left with one option. That is either a Boy or a Girl, with reference to my example, we have 2 boxes now. Naturally it becomes 50%.

Helps this helps those who still doesn't believe the solution is wrong.

Cheers!

Lightz Ong Kay Soon

Edited by Lightz

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First to prove the solution wrong:

There are only 4 possibilies of the combination, namely:

Boy, Boy 25%

Girl, Girl 25%

Boy, Girl 25%

Girl, Boy 25%

If we were to take these possibilies as a set, the chances of getting a boy with a girl is 50% whereas boy boy and girl girl are 25% each. Noted that this is where the person who comes out with the wrong solution is misleaded.

BUT look carefully, when the first option is already out. Say its a girl in this case.

Boy, Boy 25%Girl, Girl 25%

Boy, Girl 25%

Girl, Boy 25%

We will be left with only 2 combinations, it becomes:

Girl, Girl 50%

Girl, Boy 50%

What do you mean "when the first option is already out"? No where in the riddle is it stated the first kid is a girl. Your mistake was eliminating "Boy, Girl" as a possibility. The only possibility that should have been eliminated was "Boy, Boy" and what you have left should look like this:

Girl, Girl 33.3%

Boy, Girl 33.3%

Girl, Boy 33.3%

"They have two kids, one of them is a girl, what is the probability that the other kid is also a girl."

The other kid is a girl one third of the time.

If you want to see if you're right, try the following experiment with pennies that adheres to the conditions of the OP.

Lay out 100 pairs of pennies randomly. Consider pennies showing heads boys and pennies showing tails girls. Since we know the couple in the OP had at least one girl, remove all pairs that are both heads up (all boys). Count the pairs you now have. Now, how many of those pairs are both tails? Is it about half of the pairs showing or one-third?

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Hi Lightz,

Your post was long enough and detailed enough to expose where you went wrong.

Don't mean to call you out, but you were pretty dismissive of the correct analysis and those who have explained it many times.

The only problem is your premise is incorrect. Oops!

Take a look:

BUT look carefully, when the first option is already out. Say its a girl in this case.

Boy, Boy 25%

Girl, Girl 25%

Boy, Girl 25%

Girl, Boy 25%

The OP does not permit ruling out the Boy, Girl case. [only Boy, Boy.]

Quite simply then, there are three cases, not two; and the probability is 1/3, not 1/2.

OK?

You said, rather condescendingly,

Its rather ironic that I have to go into such details into explaining why the solution is wrong, because the logic is just so simple:

There are a couple rather obvious points you might have thought of before saying that:

[1] the OP states that you can assume the chance of a boy or girl birth is 50%.

What kind of a riddle would it be if the question then posed was simply: what are the odds that the 2nd child is a girl?

[2] what kind of people do you think come to this site to post and answer riddles who would not immediately say it's 50%

if that were the question being asked?

Hint: when it seems to you that a whole lot of fairly smart people have got an answer wrong,

it might be time to consider whether you got the question right.

Either way, right or wrong, respect for another's opinion is always appropriate.

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Making babies by the flip of a coin?

any birth is

5/50 though the world figures are 52/48 and that is without EVERY birth counted (particpation requires 100% of the whole world)

Though this does not affect math probability - best used for casino games!

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thoughts,

the question should be re-worded "What are the chances two girls are born" because, each in case there is a 1/2 chance there will be a boy or a girl, think about the question without having the first child, so then the probabilty of a getting a girl is 1/2. In the information given it tells us that the first child is always a girl so the probobaly of getting a girl is 1 so, 1 X 1/2 = 1/2 Chance

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Hi Bonanova

No offence. When i meant the first option is already out, simply means we already know the first answer, meaning it came out already. Meaning we do not have the advantage at this point anymore. I'll explain what i meant by losing the advantage.

Try this.

A lady gives birth, it is harder to get (both Girls) or (both Boys) than (1Girl & 1 Boy). Why?

Possibilty before first birth:

B B 25%

G G 25%

B G 25%

G B 25%

If I guess B&B. If the first birth is a G, I lose.

If I guess G&G. If the first birth is a B, I lose.

By choosing B&G, I will never lose on the first child. Important, This is where the advantage lies. B&G gives you advantage over BB or GG because 50% is greater than 25%. Note that your chances are always 50% when you choose B&G.

Before the first option is known: BG and GB are two possibilities

But once the first child is born, supposing you got a G.

You are only left with the GG or GB possibility. Now it becomes 50% = 50% (This is where we are at)

After the first option is known: BG and GB automatically becomes one possibility becos we lost the advantage of interchange.

Once we have the first outcome, we will lose the interchange advantage of the other BG or GB possibility. Yes there is definitely a higher chance of getting BG or GB at first but the advantage only occurs at one point as explained above. Once you have know you have a G, you are left with 2 combinations and the probability remains 50%.

I'll go a little further

Whatchya Gonna Do (2 Goats and a Car)

You are on a game show and there are three doors. The presenter tells you that behind one of doors there is a car and behind the other two are goats. If you pick the car you win it. After you have picked a door the presenter opens a different door with a goat behind it, he then gives you the chance to change what door you open. What should you do?

I'm pretty sure most of us read about the 2 goats a 1 car puzzle. Guess I'll just use it to further elaborate. If the host does this every week (opens a door after the contester has choosen 1 door) What i gonna do would be simply tell him "Why dont you open a door with a goat first and then ask me to come down and choose my door? There is no difference, you are gonna open a door with a goat anyway. Irregardless of I choose my door first or not, there will be no difference"

Sometimes It's just the way we see things. When we are left with 2 doors with a goat and a car behind. It's simply 50% again! *wink

Cheers

Lightz Ong Kay Soon

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Okay, I dunno why people are still debating this from over a half a year ago. O_o

Lightz, the question you are proposing as a similarity to this question is actually very different: that is the Monty Hall problem.

This is a simple question of probability, and has already been solved, proven multiple times in this topic. (And by mathematicians(although logicians might sometimes disagree?))

I won't post the answer because it's already been proved, but if people still have a problem with this I guess feel free to post?

Otherwise, I really think this is a dead topic, not worth discussing anymore.

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Okay, I dunno why people are still debating this from over a half a year ago. O_o

Lightz, the question you are proposing as a similarity to this question is actually very different: that is the Monty Hall problem.

This is a simple question of probability, and has already been solved, proven multiple times in this topic. (And by mathematicians(although logicians might sometimes disagree?))

I won't post the answer because it's already been proved, but if people still have a problem with this I guess feel free to post?

Otherwise, I really think this is a dead topic, not worth discussing anymore.

Hence why I'm surprised to see it again and again. I have my own opinion back aways. It's so so so passe.

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Hi Bonanova

No offence. When i meant the first option is already out, simply means we already know the first answer [edit - we don't], meaning it came out already. Meaning we do not have the advantage at this point anymore. I'll explain what i meant by losing the advantage.

Try this.

A lady gives birth, it is harder to get (both Girls) or (both Boys) than (1Girl & 1 Boy). Why?

Possibilty before first birth:

B B 25%

G G 25%

B G 25%

G B 25%

If I guess B&B. If the first birth is a G, I lose.

If I guess G&G. If the first birth is a B, I lose.

By choosing B&G, I will never lose on the first child. Important, This is where the advantage lies. B&G gives you advantage over BB or GG because 50% is greater than 25%. Note that your chances are always 50% when you choose B&G.

Before the first option is known: BG and GB are two possibilities

But once the first child is born, supposing you got a G. [edit: that's NOT what the OP tells us]

You are only left with the GG or GB possibility. Now it becomes 50% = 50% (This is where we are at) [edit: No, it's NOT.]

After the first option is known: [edit - it isn't] BG and GB automatically becomes one possibility becos we lost the advantage of interchange.

Once we have the first outcome [edit - we don't], we will lose the interchange advantage of the other BG or GB possibility. Yes there is definitely a higher chance of getting BG or GB at first but the advantage only occurs at one point as explained above. Once you have know you have a G, you are left with 2 combinations and the probability remains 50%.

Cheers

Lightz Ong Kay Soon

Lightz,

I appreciate your patience with me.

Now hold on, this may shock you:

If 2 + 2 = 5 then New York is a small city.

That logic is valid.

I state it because it shows what happens when the premise is false.

What happens is that valid logic can take you to a wrong conclusion.

That is, you'll have to tell us why you are insisting that the first child is a girl.

If you post again, make the first thing you do to point to the words in the OP that say the first child is a girl.

Just copy and highlight those words.

You see, the OP does nothing more than rule out the Boy-Boy possibility. Nothing more. Nothing.

If you think it rules out the Boy-Girl possibility, show where it does that.

In the Boy-Girl case, the statement "one of them is a girl" is absolutely true. There is no contradiction.

The OP - unless you can show us what we missed - DOES NOT rule out the Boy-Girl possibility.

I think that statement has been made enough times that you can't ignore it anymore.

Show us the words in the OP that rule out the Boy-Girl case, and we will agree with you.

If the B-G case is left in, then the G-G case is 1 out of 3.

And I think you agree with that.

OK?

Above all, have fun.

Oh, and leave the goat for it's own forum.

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Okay, I dunno why people are still debating this from over a half a year ago. O_o

Lightz, the question you are proposing as a similarity to this question is actually very different: that is the Monty Hall problem.

This is a simple question of probability, and has already been solved, proven multiple times in this topic. (And by mathematicians(although logicians might sometimes disagree?))

I won't post the answer because it's already been proved, but if people still have a problem with this I guess feel free to post?

Otherwise, I really think this is a dead topic, not worth discussing anymore.

ok, this is a dead topic, and not worth discussing anymore. but i spend more than 1 hour to read all the post and to recall that conditional probability, so i think that I SHOULD WRITE SOMETHING

IF the problem is the first child is a girl and what the probability the second child is also a girl, then it should be easy:

P(secondchildisagirl | firstchildisagirl) = P (firstchildisagirl n secondchildisagirl) / P(firstchildisagirl)

P (firstchildisagirl n secondchildisagirl) = 1/4

P(firstchildisagirl) = 1/2

P(secondchildisagirl | firstchildisagirl) = 1/4 / 1/2 = 1/2

but the problem is: one of the children is a girl and what the probability that the other child is also a girl,

P (otherchildisagirl | oneofthechildrenisagirl) = P (otherchildisagirl n oneofthechildrenisagirl) / P (oneofthechildrenisagirl)

P (otherchildisagirl n oneofthechildrenisagirl) = P (GG) = 1/4

P (oneofthechildrenisagirl) = P(GG)+P(BG)+P(GB)=3/4

P (otherchildisagirl | oneofthechildrenisagirl) = 1/4 / 3/4 = 1/3

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Three problems.

1. I pretty sure that this is what they call "mutually exclusive events" which bascially means that they do not affect each other. So just because the first one was a girl, it doesn't affect the probability of the second child. You can't say to someone who has a girl that your next child is more likely to be a girl.

2. even if you base it on the four choices: Girl - Girl

Girl - Boy

Boy - Girl

Boy - Boy

There is still 50% chance of the next one being a girl as you can see that you have marked out the fact that it can no longer be boy - boy, but then you also need to mark out the fact that it can no longer be boy - girl as the first child is, as you have said a GIRL!!! |So the only choices are:Girl - Girl

Girl - Boy

so 50% chance that the next is a girl, so even your own solution is flawed!!

3. If you go into biology, males have an XY sex chromosome and females have an XX chromosome, so if you do a punnet square showing the genetic cross, the possible outcomes are: XX, XX , XY, XY. so there is a 50% chance of XX (a female) and 50% chance of XY (a male).

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So just because the first one was a girl, it doesn't affect the probability of the second child. You can't say to someone who has a girl that your next child is more likely to be a girl.

Spend some time going over the posts in this thread. This riddle is not about the probability of the second child. No one is saying the next child is more likely to be a girl.

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