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# A couple decides to have 4 children. If they succeed in havi

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A couple decides to have 4 children. If they succeed in havi [#permalink]

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04 Aug 2013, 00:54
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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 havi [#permalink]

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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 havi [#permalink]

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04 Aug 2013, 01:07
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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 havi [#permalink]

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25 Dec 2013, 07:28
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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 havi [#permalink]

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10 Aug 2014, 02:40
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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 havi [#permalink]

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

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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

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

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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 havi [#permalink]

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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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A couple decides to have 4 children. If they succeed in havi [#permalink]

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04 Feb 2016, 16:50
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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A couple decides to have 4 children. If they succeed in havi [#permalink]

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04 Feb 2016, 22:35
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Dondarrion 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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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 01 Nov 2014 Posts: 16 Followers: 0 Kudos [?]: 0 [0], given: 0 Re: A couple decides to have 4 children. If they succeed in havi [#permalink] ### Show Tags 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? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7380 Location: Pune, India Followers: 2291 Kudos [?]: 15146 [0], given: 224 Re: A couple decides to have 4 children. If they succeed in havi [#permalink] ### Show Tags 08 Aug 2016, 03:22 Expert's post 1 This post was BOOKMARKED 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 _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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

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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
Re: A couple decides to have 4 children. If they succeed in havi   [#permalink] 23 Mar 2017, 08:17
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