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A couple decides to have 4 children. If they succeed in having 4 child

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A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys?

a. 3/8
b. 1/4
c. 3/16
d. 1/8
e. 1/16

OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/a-couple-dec ... 68730.html
[Reveal] Spoiler: OA

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Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink]

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New post 18 Feb 2011, 13:03
Baten80 wrote:
A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys?

a. 3/8
b. 1/4
c. 3/16
d. 1/8
e. 1/16


Probability of having BOY or GIRL is equally likely= 1/2

Having 2 girls out of 4 child can happen in C(4,2) ways

\(C^4_2*(\frac{1}{2})^4\)

\(\frac{12}{2}*\frac{1}{16}\)

\(\frac{3}{8}\)

Ans: "a"
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Baten80 wrote:
A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys?

a. 3/8
b. 1/4
c. 3/16
d. 1/8
e. 1/16


\(P(GGBB)=\frac{4!}{2!2!}*(\frac{1}{2})^4=\frac{3}{8}\), we should mulitply by \(\frac{4!}{2!2!}\) as the scenario GGBB can occur in several ways: GGBB, BBGG, GBGB, ... # of scenarios possible is # of permutations of 4 letters GGBB out of which 2 G's and 2 B's are identical and equals to \(\frac{4!}{2!2!}\).

Answer: A.
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Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink]

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A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, What is the probability that they will have exactly 2 girls and 2 boys?

(A) 3/8
(B) 1/4
(C) 3/16
(D) 1/8
(E) 1/16



I see that we can use combination rule here [4! / 2! (2!)]/2^4. What if the question was 1 girls and 2 boys? What will be the nominator?

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Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink]

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New post 04 Aug 2013, 00:59
ksung84 wrote:
A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, What is the probability that they will have exactly 2 girls and 2 boys?

(A) 3/8
(B) 1/4
(C) 3/16
(D) 1/8
(E) 1/16



I see that we can use combination rule here [4! / 2! (2!)]/2^4. What if the question was 1 girls and 2 boys? What will be the nominator?


If that was the case, the question would be similar to arranging B,B,G -->\(\frac{3!}{2!}\).
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Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink]

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Since there is an equal probability of having a boy or a girl, Any possible combination of 4 children has a probability of 1/16. Now there are 4!/2!*2! ways of having exactly 2 boys and 2 girls... which is 6. So total probability of that event happening is 6*1/16 = 3/8 = A

The answer would not change for 2 boys and 1 girl

Which will be 3*1/8 = 3/8
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Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink]

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Since probability that the couple will have a boy or a girl is 1/2, likelihood of having 2 boys and 2 girls is 1/2* 1/2 * 1/2 * 1/2 = 1/16

Now, 2 boys out of 4 can be chosen in 4c2 i.e 6 ways. Choosing 2 boys out of 4 leaves 2 girls, hence no additional selection of girls is required.

Finally, probability of choosing exactly 2 boys and 2 girls is 1/16 * 6 = 3/8.

Hope that helps.
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Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink]

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By fundamental counting principle,
Total No of out comes: 2^4 = 16

Total Desired outcomes: No of ways to arrange BBGG by MISSISSIPPI rule= 4!/(2!*2!) = 6

Probability is 6/16 = 3/8
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Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink]

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New post 24 Nov 2014, 14:26
Referring to the question: What if the question was 1 girls and 2 boys? What will be the nominator?

You can easily verify this by using the same methodology. There are _ _ _ --> 3 possible spots, 1 for every child. Each of these spots has only 2 possible outcomes occurring with equal likelihood (1/2).

This leaves us with a probability of (1/2)^3 =1/8 for every single possible outcome.

The next thing to ask yourself is how many of all possible cases include 2 boys (B) and 1 girl (G). The answer is 3 and turns out to be very intuitive if you start to write down a few possible outcomes.

GBB
BGB
BBG

All other scenarios will either include 2 girls and 1 boy, 3 boys or 3 girls.

Final step is to multiply the probability of each individual outcome with the number of outcomes that satisfy the condition we are looking for. --> 3x(1/8)=3/8

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New post 24 Nov 2014, 22:21
One case is 1/2*1/2*1/2*1/2=1/16

counting symmetry to know how many cases we can have, so 4!/2!*2!=6

1/16*6=6/16=3/8

A

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Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink]

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New post 16 Jul 2015, 20:06
Why are we ordering here? I mean, I would think BBGG to be the same as BGBG - since the question asks about exactly 2 girls and 2 boys, irrespective of any order of delivery.
We have 5 possibilities: 4 boys, 3 boys and 1 girl, 2 boys and 2 girls, 1 boy and 3 girls, and finally all 4 girls.
So shouldn't the probability be 1/5? too simplistic - I know. but where am I wrong?

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Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink]

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New post 16 Jul 2015, 21:14
ksung84 wrote:
A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, What is the probability that they will have exactly 2 girls and 2 boys?

(A) 3/8
(B) 1/4
(C) 3/16
(D) 1/8
(E) 1/16



I see that we can use combination rule here [4! / 2! (2!)]/2^4. What if the question was 1 girls and 2 boys? What will be the nominator?


Sample space = 2^4 = 16.

Favourable events = {bbgg}, {bgbg}, {bggb}, {ggbb}, {gbgb}, {gbbg}.

Probability = 6/16 = 3/8. Ans (A).
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Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink]

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New post 04 Feb 2016, 16:50
donkadsw wrote:
Why are we ordering here? I mean, I would think BBGG to be the same as BGBG - since the question asks about exactly 2 girls and 2 boys, irrespective of any order of delivery.
We have 5 possibilities: 4 boys, 3 boys and 1 girl, 2 boys and 2 girls, 1 boy and 3 girls, and finally all 4 girls.
So shouldn't the probability be 1/5? too simplistic - I know. but where am I wrong?


Bump seeking an answer to the question above:

I got 1/5 as well.

BBBB
BBBG
BBGG
GGGB
GGGG

1/5 outcomes

The question doesn't say anything about the order. How do we know when order matters?

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Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink]

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New post 04 Feb 2016, 22:35
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Dondarrion wrote:
donkadsw wrote:
Why are we ordering here? I mean, I would think BBGG to be the same as BGBG - since the question asks about exactly 2 girls and 2 boys, irrespective of any order of delivery.
We have 5 possibilities: 4 boys, 3 boys and 1 girl, 2 boys and 2 girls, 1 boy and 3 girls, and finally all 4 girls.
So shouldn't the probability be 1/5? too simplistic - I know. but where am I wrong?


Bump seeking an answer to the question above:

I got 1/5 as well.

BBBB
BBBG
BBGG
GGGB
GGGG

1/5 outcomes

The question doesn't say anything about the order. How do we know when order matters?


In this case we are saying that the probability of BGGG is the same as the probability of BBGG. But that is not so.

You can have 1 boy and 3 girls in 4 ways: BGGG, GBGG, GGBG, GGGB
But you can have 2 boys and 2 girls in 6 ways: BBGG, BGGB, GGBB, BGBG, GBGB, GBBG

So the probability depends on the number of ways in which you can get 2 boys and 2 girls.

Think of it this way: if you throw two dice, is the probability of getting a sum of 2 same as the probability of getting a sum of 8? No.
For sum fo 2, you must get 1 + 1 only.
For sum of 8, you could get 4 + 4 or 3 + 5 or 2 + 6 etc.
So probability of getting sum of 8 would be higher.

In the same way, the order matters in this question.
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New post 06 Aug 2016, 15:17
Hi guys, I don't quite understand why we square 4 instead of factorial of 4. Could you please help?

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ManonZ wrote:
Hi guys, I don't quite understand why we square 4 instead of factorial of 4. Could you please help?



Do you mean in the total number of cases? If yes, then it is actually 2^4.
First child can happen in 2 ways (boy or girl).
Second, third and fourth kids can also happen in 2 ways.
Total number of ways = 2*2*2*2 = 16
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Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink]

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New post 23 Mar 2017, 08:17
probability question: total number of outcomes is 2*2*2*2=16
the # of desired outcomes can be found 2C4, or 4!/2!2!=6
6/16=3/8
Answer is A

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New post 31 May 2017, 19:33
ksung84 wrote:
A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, What is the probability that they will have exactly 2 girls and 2 boys?

(A) 3/8
(B) 1/4
(C) 3/16
(D) 1/8
(E) 1/16



I see that we can use combination rule here [4! / 2! (2!)]/2^4. What if the question was 1 girls and 2 boys? What will be the nominator?


1.This requires ordering and repetition. So the formula is n^r for the total number of cases. By ordering I mean bbgg is different from ggbb and repetition happens because b or g can happen more than once.
2. Favorable cases are : bbgg, bgbg, bggb, ggbb, gbgb, gbbg
Probability is therefore 6/16=3/8
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Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink]

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New post 02 Jun 2017, 10:44
ksung84 wrote:
A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, What is the probability that they will have exactly 2 girls and 2 boys?

(A) 3/8
(B) 1/4
(C) 3/16
(D) 1/8
(E) 1/16



I see that we can use combination rule here [4! / 2! (2!)]/2^4. What if the question was 1 girls and 2 boys? What will be the nominator?


We can solve this problem in a couple of ways.

First, realize that there are 2^4 = 16 total possibilities (boy/girl = 2 options * 4 instances).

Next, you can realize that there is 1 way to have all boys and 1 way to have all girls. So the total number of invalid options is 2.

Then, if you have 1 boy and 3 girls or if you have 1 girl and 3 boys, then you the chances of these outcomes occurring should be equal. And there are 4 ways to have 1 boy, so there must be 4 ways to have one girl.

The total number of invalid cases is 2+4+4 = 10 out of 16 total options. So the number we care about is the remaining number of options with 2 boys and 2 girls = 16-10 = 6.

6/16 = 3/8.

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Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink]

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New post 19 Jun 2017, 02:56
I did it the binomial way.
Proba of having a girl is same as proba of having a boy, which is 1/2.
Proba of having exactly 2 girls and 2 boys is given by 4C2 * (1/2)^2 * (1/2)^2 = 6/16 or 3/8.

Answer is A

The question would be much more interesting if proba of having a girl was 1/3 for instance, and we were asked to find the proba of having exactly 3 girls and 1 boy. In this case the answer would have been: 4C3 * (1/3)^3 * (2/3)^1 = 8/81.

Binomial Proba is fast and easy...
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Re: A couple decides to have 4 children. If they succeed in having 4 child   [#permalink] 19 Jun 2017, 02:56

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