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A couple decides to have 4 children. If they succeed in havi [#permalink]
03 Aug 2013, 23:54

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Difficulty:

65% (hard)

Question Stats:

55% (02:14) correct
45% (00:54) wrong based on 189 sessions

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]
03 Aug 2013, 23:59

Expert's post

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]
04 Aug 2013, 00:07

4

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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 _________________

You've been walking the ocean's edge, holding up your robes to keep them dry. You must dive naked under, and deeper under, a thousand times deeper! - Rumi

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

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