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

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03 Aug 2013, 23:54

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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?

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

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03 Aug 2013, 23:59

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 -->\(\frac{3!}{2!}\).
_________________

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

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04 Aug 2013, 00:07

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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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24 Nov 2014, 13:26

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

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

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16 Jul 2015, 19:06

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?

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

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16 Jul 2015, 20:14

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?

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

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04 Feb 2016, 15:50

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?

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.
_________________

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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