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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] New post 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?
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Re: A couple decides to have 4 children. If they succeed in havi [#permalink] New post 03 Aug 2013, 23:59
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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] New post 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] New post 25 Dec 2013, 06:28
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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] New post 10 Aug 2014, 01:40
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] New post 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
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Re: A couple decides to have 4 children. If they succeed in havi [#permalink] New post 24 Nov 2014, 21:21
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] New post 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?
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Re: A couple decides to have 4 children. If they succeed in havi [#permalink] New post 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?


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 havi   [#permalink] 16 Jul 2015, 20:14
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