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What is the probability of bearing two boys from four born

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Senior Manager
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What is the probability of bearing two boys from four born [#permalink] New post 06 Aug 2007, 09:19
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A
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D
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What is the probability of bearing two boys from four born babies?
(A) 1/5
(B) 1/8
(C) 1/2
(D) 3/8
(E) 6/11
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 [#permalink] New post 06 Aug 2007, 09:38
I believe it goes like this:

total number of outcomes = 2*2*2*2 = 16
total number of outcomes with 2 Boys = 4 choose 2 = 4! / (2!*2!) = 6

answer is 6/16=3/8 (choice D)

Hope this helps!
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 [#permalink] New post 06 Aug 2007, 10:06
trancing wrote:
I believe it goes like this:

total number of outcomes = 2*2*2*2 = 16
total number of outcomes with 2 Boys = 4 choose 2 = 4! / (2!*2!) = 6

answer is 6/16=3/8 (choice D)

Hope this helps!


Please, explain how did you get total number of outcomes?
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 [#permalink] New post 06 Aug 2007, 10:23
4 babies where each one of them is either a boy or girl

2 choices for baby 1 * 2 choices for baby 2 * 2 choices for baby 3 * 2 choices for baby 4 = 2 ^4 = 16 total outcomes.

Btw, what answer does the OA give?
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 [#permalink] New post 09 Aug 2007, 08:39
trancing wrote:
I believe it goes like this:

total number of outcomes = 2*2*2*2 = 16
total number of outcomes with 2 Boys = 4 choose 2 = 4! / (2!*2!) = 6

answer is 6/16=3/8 (choice D)

Hope this helps!


Good explanation..

Another way I will at it is :

Total outcome is 16.

2 boys off 4 kids, I have following possibilities:

BBGG
BGBG
BGGB
GBGB
GGBB
GBBG

So, 6 possibilities out 16.
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 [#permalink] New post 11 Aug 2007, 23:27
Thanks :)
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Re: PS question (probability, another one) [#permalink] New post 12 Aug 2007, 00:13
Vlad77 wrote:
What is the probability of bearing two boys from four born babies?
(A) 1/5
(B) 1/8
(C) 1/2
(D) 3/8
(E) 6/11


got (D) 3/8

2 boys: 4C2 = 4*3 / 2*1 = 6
total: 2^4 = 2*2*2*2 = 16

F/T = 6/16 = 3/8
Re: PS question (probability, another one)   [#permalink] 12 Aug 2007, 00:13
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