A simple logic problem

[timotinbowen]

I knew I saw this before............Here is the Rundown
As the answer could be Both 2/3 or 1/2

Supposing that we randomly pick a _child_ from a two-child family. We
see that he is a boy, and want to find out whether his sibling is a
brother or a sister. (For example, from all the children of two-child
families, we select a child at random who happens to be a boy.) In
this case, an unambiguous statement of the question could be:

From the set of all families with two children, a child is
selected at random and is found to be a boy. What is the
probability that the other child of the family is a girl?

Note that here we have a pool of kids (all of whom are from two-child
families) and we're pulling one kid out of the pool. This is like the
problem you're talking about. The child selected could have an older
brother, an older sister, a younger brother or a younger sister.

Let's look at the possible combinations of two children. We'll use B
for Boy and G for girl, and for each combination we'll list the older
child first, so GB means older sister while BG means younger sister.
There are 4 possible combinations:

{BB, BG, GB, GG}

From these possible combinations, we can eliminate the GG combination
since we know that one child is a boy. The three remaining possible
combinations are:

{BB, BG, GB}

In these combinations there are four boys, of whom we have chosen one.
Let's identify them from left to right as B1, B2, B3 and B4. So we
have:

{B1B2, B3G, GB4}

Of these four boys, only B3 and B4 have a sister, so our chance of
randomly picking one of these boys is 2 in 4, and the probability is
1/2 - as you have indicated.


But now let's look at a different way of selecting the "boy" in the
problem. Suppose we randomly choose the two-child _family_ first. Once
the family has been selected, we determine that at least one child is
a boy. (For example, from all the mothers with two children, we select
one and ask her whether she has at least one son.) In this case, an
unambiguous statement of the question could be:

From the set of all families with two children, a family is
selected at random and is found to have a boy. What is the
probability that the other child of the family is a girl?

Note that here we have a pool of families (all of whom are two-child
families) and we're pulling one family out of the pool. Once we've
selected the family, we determine that there is, in fact, at least one
boy.

Since we're told that one child (we don't know which) is a boy, we can
eliminate the GG combination. Thus, our remaining possible
combinations are:

{BB, BG, GB}

Each of these combinations is still equally likely because we picked
one of the four families.

Now we want to count the combinations in which the "other" child is a
girl. There are two such combinations: BG and GB.

Since there are three combinations of possible families, and in two of
them one child is a girl, the probability is 2/3.

The question posed clearly fits into one of the two categories. I provided the information that there is a 2-child set, of which at least 1 is a boy. The reasoning behind the answers were interesting, and I wish you would've waited before posting this (unless of course it's an original), but the line of thinking in each case is correct (and I'll leave it up to everyone else to determine which question was being asked, and therefore what the correct answer is).

 

[quantumleap]

"There are 4 possible combinations:

{BB, BG, GB, GG}"

Once it has been determined that one of the possibilities is B then the 4 possibilities are no longer valid. The knowledge of part of the outcome invalidates the original possibilities.

Therefore only BB, BG or GB can be considered. Since GB and BG is the same there can only be 1 of 2 possible remaining outcomes. 50%.

It's like asking someone what the probability that a coin will end up heads if flipped 100 times. The answer is 50%. Then, the coin is flipped and it turns up heads. What is the probability that the coin will end up heads in 99 times. It's not 49 times out of 99 since the first one ended up heads. It's still 50%.

 

[timotinbowen]

"There are 4 possible combinations:

{BB, BG, GB, GG}"

Once it has been determined that one of the possibilities is B then the 4 possibilities are no longer valid. The knowledge of part of the outcome invalidates the original possibilities.

Therefore only BB, BG or GB can be considered. Since GB and BG is the same there can only be 1 of 2 possible remaining outcomes. 50%.

It's like asking someone what the probability that a coin will end up heads if flipped 100 times. The answer is 50%. Then, the coin is flipped and it turns up heads. What is the probability that the coin will end up heads in 99 times. It's not 49 times out of 99 since the first one ended up heads. It's still 50%.

Once we get added information, we CAN eliminate one possibility, while keeping all other selection sets. It is important to realize that the problem states that one of the children is a boy, not that a particular one is a boy. GB and BG give the same amount of boys, but they are permutations which each account for (after the GG selection is eliminated) 1/3 of the possible combinations.

Another way of coming to the answer (2/3), which is slightly trickier, is as follows:
We know that the probability of 2 children being 1 boy and 1 girl is 50% (because the first will either be B or G, and then it is 50% whether the second is the other gender). We also know that the probability of having at least 1 Girl is 3/4: (1-(.5*.5)). We now need to find G so that 3/4 * G = 1/2. Solving this simple equation gives G = 2/3.

It is important to notice the differences between this problem and your coin example. This is a problem of partial information, whereas your problem has no information. Further, we are not told that the FIRST child is a boy, just that there is a boy in the set (this is critical). To make your coin problem the same we could ask:

A coin is flipped twice and comes up Heads at least once. What is the probability that it came up Heads once and Tails once?

 

[goodcall]

reminds me of that 3 aces question

 

[seriousaddiction]

who was on top?

 

[festeringZit]

A man tells you he has two children. He then starts talking about his son. He does not tell you whether the son is the oldest child or the youngest child. What is the probability that his other child is a girl?

The answer as the problem as worded is 2/3.

There are 4 possibilities:

BB, BG, GB, GG

The fact that he mentions a boy rules out the last one, so now we
have three equally likely options: BB, BG, GB

We don't know which boy he is talking about in the BB case, so there
is 1/6 chance he is talking about the first boy in the BB case, and 1/6 chance he is talking about the second boy in the BB case.

1/3 BB
1/3 BG
1/3 GB

Out of these options, the answer is clearly 2/3 chance that the other
child is a girl.

 

[MattRain]

1/3 BB
1/3 BG
1/3 GB

BG is the same as GB. If you have 2 kids, you either have 2 girls, 2 boys, or one of each. 3 possibilities, not 4.

Answer is (roughly) 50%. You're being asked what the chances are that a random person is male or female.

 

[festeringZit]

BG is the same as GB. If you have 2 kids, you either have 2 girls, 2 boys, or one of each. 3 possibilities, not 4.

Answer is (roughly) 50%. You're being asked what the chances are that a random person is male or female.

Sorry, but you are 100% wrong.

Consider a computer program that flips two coins, and
prints out the results 1,000,000 times.

HH
HT
TH
TT
HH
HT
TH
TT
... -> 1,000,000

In looking at the list of results you are told that you must remove
the entries that contain TT. Now, given the rest of the results,
what % of the results will contain a T?

2/3.

It's the same problem.

Now if the problem were worded: "the man talks about his youngest child who is a boy, what
is the probability the other child is a girl?" The answer would be 1/2.

 

[MattRain]

You're being asked what the chances are that a random person is female.

Fixed my own post.

 

[MattRain]

Sorry, but you are 100% wrong.

Consider a computer program that flips two coins, and
prints out the results 1,000,000 times.

HH
HT
TH
TT
HH
HT
TH
TT
... -> 1,000,000

In looking at the list of results you are told that you must remove
the entries that contain TT. Now, given the rest of the results,
what % of the results will contain a T?

2/3.

It's the same problem.

Explain to me the difference between having a girl and a boy, and having a boy and a girl.

There is none. HT and TH are duplicates.

 

[festeringZit]

Explain to me the difference between having a girl and a boy, and having a boy and a girl.

There is none. HT and TH are duplicates.

The key is that they are two different possibilities that each have
an associated probability assigned to them. Keep looking at the
coin flip computer simulation, and I think the light will come on.

 

[MattRain]

The light is on, thank you very much. The problem is so simple, I don't know why you guys complicate it with procedural BS.

If you wanna consider BG and GB separately, that means you eliminate BOTH GG and GB when the guy tells you he has Bx.

1/2. Come on people.

 

[Betallsports]

The question posed clearly fits into one of the two categories. I provided the information that there is a 2-child set, of which at least 1 is a boy. The reasoning behind the answers were interesting, and I wish you would've waited before posting this (unless of course it's an original), but the line of thinking in each case is correct (and I'll leave it up to everyone else to determine which question was being asked, and therefore what the correct answer is).

Sorry about that, just remembered that I saw it before.
And Yes, it is the Original from 2000 or 2001

But a Great Question :toast:

 

[festeringZit]

The light is on, thank you very much. The problem is so simple, I don't know why you guys complicate it with procedural BS.

If you wanna consider BG and GB separately, that means you eliminate BOTH GG and GB when the guy tells you he has Bx.

1/2. Come on people.

Matt,

I'll bet you $100, or any other amount of your choice
that you are wrong. OK?

 

[Betallsports]

Explain to me the difference between having a girl and a boy, and having a boy and a girl.

There is none. HT and TH are duplicates.

Sorry to say, there is a Difference.

Are you a Craps Bettor ? How many possibilities to make a 6 or an 8 ??

 

[MattRain]

I guess I'm missing something. Let me go through the whole thread.

 

[Geoff101]

guys guys guys...the probability of the second child's sex has absolutely no relevance to the first child's sex....in other words those computing:
BG
GB
BB
GG
are WRONG....there is NO possible way the first is a G, because he tells you the first child is a Boy...
Let's pose this question and see what responses we get:
The father has 1400 children, 1399 of them are boys, what are the odds that the 1400th is a girl?

50%

 

[festeringZit]

One can also use Bayes' Theorem to figure this out.

<dl><dd>

</dd></dl>A = event that there is one boy and one girl
B = event that there is at least one boy


2/4 / 3/4 = 8/12 = 2/3

 

[MattRain]

I see how 2/3 would be the answer if you use craps or coins... but in the real world, what kind of parent wishes they had a boy and a girl instead of a girl and a boy?

Lit major here, sorry for rejecting your rational mathspeak. :grandmais

1/2. Vive la resistance.

 

[festeringZit]

guys guys guys...the probability of the second child's sex has absolutely no relevance to the first child's sex....in other words those computing:
BG
GB
BB
GG
are WRONG....there is NO possible way the first is a G, because he tells you the first child is a Boy...
Let's pose this question and see what responses we get:
The father has 1400 children, 1399 of them are boys, what are the odds that the 1400th is a girl?

50%

100% wrong. And, he did not tell you the *first* child is a boy.

sigh.