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A man and a woman have two children each. At least one of the woman's children is a boy, and the man's oldest child is a son.

What is the probability that the woman's child not mentioned is a boy? And the man's? Are they even different?

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i have no ability for probability but if the man and woman do not have children in common then I think the answer is 50% that the other child is a boy for each of them....if they have the same children then...well I have no idea. heh

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Sorry if it wasn't clear. Their children are not the same and they are not husband/wife or any sort of relation.

And nope that's not the correct answer. :)

Edited by Noct

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then with no mind for the math of probability I have to stick with 1/2 chance the other is a boy. Either it is or it isn't. Someone will show me the math I am sure.

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This has been posted before, slightly modified.

The answers are:

For the woman, the equally likely cases are [eldest first] BB BG GB.

So the chances of BB are 1 in 3.

For the man, the equally likely cases are [eldest first] BB, BG

So the chances of BB are 1 in 2.

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This has been posted before, slightly modified.

The answers are:

For the woman, the equally likely cases are [eldest first] BB BG GB.

So the chances of BB are 1 in 3.

For the man, the equally likely cases are [eldest first] BB, BG

So the chances of BB are 1 in 2.

correct ... there was a loooong discussion ... check One Girl - One Boy

before you continue in this thread, have a look at the original variation

thread open

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correct ... there was a loooong discussion ... check One Girl - One Boy

before you continue in this thread, have a look at the original variation

thread open

That's more than stupid.Doesn't matter the order of the children.No matter if the first born child is girl or boy the chances are always 1/2 for both genders.Even if you have 10 girls the chance the 11th child to be boy still is 1/2.

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B is known boy, b or g for unknown child

WOMAN: Bb, bB Bg gB; probability of 2 boys is 2/4 or 1/2

MAN: Bg Bb

probabiliy of 2 boys is 1/2

This problem is written in probability books as "a king has one living sibling. What is the probability that the sibling is a boy."

Here the possibilities are :Bb Bg or gB, so the probability is 1/3. bB is not an option, as the oldest living son is the king. Sisters could be older, as it is the oldest living son that gets the kingdom.

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That's more than stupid.Doesn't matter the order of the children.No matter if the first born child is girl or boy the chances are always 1/2 for both genders.Even if you have 10 girls the chance the 11th child to be boy still is 1/2.

Not quite. the order doesn't matter in the sense of we didn't specify the order, but it does effect the probability.

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B is known boy, b or g for unknown child

WOMAN: Bb, bB Bg gB; probability of 2 boys is 2/4 or 1/2

MAN: Bg Bb

probabiliy of 2 boys is 1/2

This problem is written in probability books as "a king has one living sibling. What is the probability that the sibling is a boy."

Here the possibilities are :Bb Bg or gB, so the probability is 1/3. bB is not an option, as the oldest living son is the king. Sisters could be older, as it is the oldest living son that gets the kingdom.

You made a mistake. Bb and bB are the same thing aren't they? That should change your answers to the correct ones :)

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A man and a woman have two children each. At least one of the woman's children is a boy, and the man's oldest child is a son.

What is the probability that the woman's child not mentioned is a boy? And the man's? Are they even different?

If you consider the unmentioned child...then it is totally independent of the other child...So, the probability is always 1/2..

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If you consider the unmentioned child...then it is totally independent of the other child...So, the probability is always 1/2..

It is dependent because which don't know which child is the boy, so we have to take all possibilities into account, and it affects the probability.

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It is dependent because which don't know which child is the boy, so we have to take all possibilities into account, and it affects the probability.

You said "not mentioned". You mentioned at least one of the child is boy...means you mentioned him...then how come the unmentioned child be conditional one the mentioned one...they r independent events...same for the man...

You might think of the sample space...b/g, g/b or b/b....but the problem is..if you mention ONE of them is a boy...then if the OTHER is girl then b/g and g/b carries same meaning...so the probability is 0.5...

Dont cofuse it with the coin toss problem with conditional probability..

Edited by storm

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You said "not mentioned". You mentioned at least one of the child is boy...means you mentioned him...then how come the unmentioned child be conditional one the mentioned one...they r independent events...same for the man...

You might think of the sample space...b/g, g/b or b/b....but the problem is..if you mention ONE of them is a boy...then if the OTHER is girl then b/g and g/b carries same meaning...so the probability is 0.5...

Dont cofuse it with the coin toss problem with conditional probability..

They carry essentially the same meaning, but the probability is greater. You have a 2/3 probability that the other child is a girl given one is a boy.

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Technically the probability is 1/2. There is Girl-Boy and there is Girl-Girl. If you counted Boy-Girl, it would no longer be a probability puzzle, it would be a combination puzzle. When working probability Girl-Boy and Boy-Girl is the same thing.

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The woman's probability is still 1/2 and here's why:

You know she has a son. Here are the four possibilities:

1. He has an older brother

2. He has a younger brother

3. He has an older sister

4. He has a younger sister

That equals 2/4 or 1/2

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Technically the probability is 1/2.

There is Girl-Boy and there is Girl-Girl.

If you counted Boy-Girl, it would no longer be a probability puzzle, it would be a combination puzzle.

When working probability Girl-Boy and Boy-Girl is the same thing.

Ask yourself: of the families that have two children, what are the odds that there is a boy and a girl?

I would say there are four equally likely cases:

BB

BG

GB

GG

and two of them are favorable, so the odds are 50%

You would say,

When working probability Girl-Boy and Boy-Girl is the same thing.

so there are 3 cases - two boys, two girls, and one each.

Three cases, and one of them is favorable, so the odds are 1/3.

Are you still comfortable with your reasoning?

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The woman's probability is still 1/2 and here's why:

You know she has a son. Here are the four possibilities:

1. He has an older brother

2. He has a younger brother

3. He has an older sister

4. He has a younger sister

That equals 2/4 or 1/2

Think about your case 1 and case 2.

You are counting the same case [she has two boys] twice.

If "he" has a younger brother [case 2], then his younger brother [who has equal right to be called "he"] has an older brother [= case 1].

Think about how the OP differs from this question:

[this is the question that you answered; try to see how it differs from the OP]

The woman has a son, named John.

John has a sibling.

What are the odds that John's sibling is a girl?

Take your four cases, substitute John for he, and you get the correct answer.

-----------------

Step 1. Enumerate the set of all equally likely outcomes

Step 2. Count the favorable outcomes.

Sept 3. Take the ratio.

Everyone "gets" this procedure, but Step 1 must be done with a sharp eye for all and an unbiased eye for equally likely.

shaunarenee combined two equally likely cases BG and GB into one [twice as likely] case.

You took one of the equally likely cases BB and made it into two.

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Think about your case 1 and case 2.

You are counting the same case [she has two boys] twice.

If "he" has a younger brother [case 2], then his younger brother [who has equal right to be called "he"] has an older brother [= case 1].

Think about how the OP differs from this question:

The woman has a son, named John.

John has a sibling.

What are the odds that John's sibling is a girl?

Take your four cases, substitute John for he, and you get the correct answer.

-----------------

Step 1. Enumerate the set of all equally likely outcomes

Step 2. Count the favorable outcomes.

Sept 3. Take the ratio.

Everyone "gets" this procedure, but Step 1 must be done with a sharp and unbiased eye for all and for equally likely.

shaunarenee combined two of them BG and GB into one [twice as likely] case.

You took one of them BB and counted it twice.

I don't see how knowing his name makes any difference to the question. You still don't know if the son is the first or second child, so both of my cases still work.

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I don't see how knowing his name makes any difference to the question. You still don't know if the son is the first or second child, so both of my cases still work.

But they are the same thing. Either way it's two boys.

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I read the first couple and last couple pages of the other thread, and I saw some good explanations which doubters should consider, but I'll toss in my two cents anyhow ...

It's a challenge to determine from the wording of the problem what the dependencies are. It requires careful consideration of what information you actually know. Consider some examples:

1. A woman has two children, at least one of whom is a boy. What is the probability she has two sons?

In this case the question clearly requires that we work with sets, and the possible outcomes are dependent on each other. It's easy to see the answer is 1/3.

2. A woman has three children. The oldest is a boy, and at least two are boys. What is the probability that all three are boys?

What do we know? We know the gender of the oldest, so the gender of the younger two is independent of that fact. Therefore, this is actually the same as the previous question. 1/3

3. A woman has two children. The oldest is a boy. What is the probability that the younger is a boy?

Again, clearly no dependence, so the answer is the natural 50%

4. A woman has two children. One is a boy. What is the probability that the other is a boy?

Is this the same as question 1? Actually it is. It appears to be asking about the probability of the gender of a single child, and thus the intuitive answer is 50%. It seems reasonable that the genders of the two children are independent, but look carefully. The gender of the child which is not in question has not been stated. This isn't a matter of ambiguous language. It is simply incorrect to say that 50% answers the question, because the gender of both children is still unknown. You have to solve this in the same manner you would question 1.

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But they are the same thing. Either way it's two boys.

Guess what? That argument still makes my point...if you're counting both of those as "two boys" then boy/girl and girl/boy are both "one boy, one girl" and therefore, it is still 1/2 probability

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A man and a woman have two children each. At least one of the woman's children is a boy, and the man's oldest child is a son.

What is the probability that the woman's child not mentioned is a boy? And the man's? Are they even different?

Are they even different?

Surely this needs some definition, are there four children between them or three or two - this may have an effect on the answers, but nothing in the OP says they are or are not related, therfore can there be a third or fourth and what excludes there being a chance that the other/s if they exist are girls.

I believe the conundrum can not be solved with the information given, only discussed.

I am tempted to say that it has no conclusion. -actually i am saying it!

Brady bunch theory or regular marriage/monogamous relationship - it's open.

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Are they even different?

Surely this needs some definition, are there four children between them or three or two - this may have an effect on the answers, but nothing in the OP says they are or are not related, therfore can there be a third or fourth and what excludes there being a chance that the other/s if they exist are girls.

I believe the conundrum can not be solved with the information given, only discussed.

I am tempted to say that it has no conclusion. -actually i am saying it!

Brady bunch theory or regular marriage/monogamous relationship - it's open.

Check post #3 regarding this

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Guess what? That argument still makes my point...if you're counting both of those as "two boys" then boy/girl and girl/boy are both "one boy, one girl" and therefore, it is still 1/2 probability

If i have 2 children, what is the probability that they are both boys? By your logic it would be 1/3, but that's clearly not the case. If you say the answer is 1/3 to the question just mentioned, then I'm sorry for you.

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