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A couple want to have four babies, for each baby, 50% are

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A couple want to have four babies, for each baby, 50% are [#permalink] New post 15 Feb 2005, 23:48
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A couple want to have four babies, for each baby, 50% are male, 50% are female. Ask for the possibility of two boys and two girls?
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 [#permalink] New post 16 Feb 2005, 00:54
I think your question is incomplete.....please specify the number of babies
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 [#permalink] New post 16 Feb 2005, 00:54
I assume what you want to ask, is what is the probability of getting 2 girls and 2 boys, and for each child, there is a 50% possibility for either gender.

So what you need is:

BOY AND BOY AND GIRL AND GIRL (independent event)

So P= 1/2*1/2*1/2*1/2 = 1/16
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 [#permalink] New post 16 Feb 2005, 01:00
antmavel, it's been a while since the last time we communicated.
just for a spot of cheekiness, it's "same method as ywilfred" instead of "same method than" (GMAT SC) :lol:
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 [#permalink] New post 16 Feb 2005, 01:15
ywilfred wrote:
antmavel, it's been a while since the last time we communicated.
just for a spot of cheekiness, it's "same method as ywilfred" instead of "same method than" (GMAT SC) :lol:


My verbal skills are terrible, i am really a poor french guy :roll:
That's why I only hunt in the Quant forum :twisted:
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 [#permalink] New post 21 Feb 2005, 00:47
shouldnt it be 3/8

total outcomes... 2*2*2*2 = 16

Our Outcomes
BBGG
BGBG
BGGB
GBBG
GBGB
GGBB
= 6

6/16 = 3/8 ??? whats the OA ?
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 [#permalink] New post 21 Feb 2005, 05:25
Prob is (B+G)^4=(1/2+1/2)^4 and the possible outcomes are:
bbbb=1/16
bggg=4/16
bbgg=6/16
bbbg=4/16
gggg=1/16
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 [#permalink] New post 09 Mar 2005, 07:25
Antmavel wrote:
My verbal skills are terrible, i am really a poor french guy :roll:
That's why I only hunt in the Quant forum :twisted:


Huh? If this is the case you should really hunt the Verbal forum more Antmavel. ;)
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 [#permalink] New post 09 Mar 2005, 07:27
Please read this thread and see if it gives you some light. :)
http://www.gmatclub.com/phpbb/viewtopic.php?t=14548
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 [#permalink] New post 09 Mar 2005, 07:56
Read this and see if you could do it again, vprabhala.
http://www.gmatclub.com/phpbb/viewtopic ... 9947#89947
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 [#permalink] New post 09 Mar 2005, 08:09
but it doesn't say atleast. it says.. what is the probability of getting 2 boys and 2 girls right Honghu?
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 [#permalink] New post 09 Mar 2005, 08:11
there is a difference between atleast and exactly i guess.

following the method.

1-none of them are boys(means all girls)- one of them is a boy(means 3 girls)
should it be giving 2 boys and 2 girls.

that would be 13/16
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 [#permalink] New post 09 Mar 2005, 08:25
I believe it's 3/8. Use the distribution formula

c(n,k)*p^k*(1-p)^(n-k).

Thanks HongHu. The theory and examples you've provided in the post above are real useful.
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 [#permalink] New post 09 Mar 2005, 08:42
4C2*(1/2)^2*(1/2)^2=3/8
I still don't get hte difference between exactly, atleast etc..
can Honghu please explain..
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 [#permalink] New post 09 Mar 2005, 08:53
vprabhala wrote:
4C2*(1/2)^2*(1/2)^2=3/8
I still don't get hte difference between exactly, atleast etc..
can Honghu please explain..


exactly means EXACTLY 2 boys and 2 girls in different combinations => bbgg or bggb or bggb, etc..

atleast 1 boy means bggg or bbgg or bbbg or bbbb,... => 4c1*(1/2)^4 + 4c2*(1/2)^4 + 4c3*(1/2)^4 + 4c4*(1/2)^4

atleast 1 boy and 1 girl bggg or bbgg or gggb,... => same way as as it is above
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 [#permalink] New post 09 Mar 2005, 21:07
3/8*3/8 for me ->prob is 9/64 to get exactly 2 boys and 2 girls

what's the OA ? Hong Hu ?
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 [#permalink] New post 09 Mar 2005, 21:45
it is 3/8.
2B and 2G = (4c2+4c2) = 6
total possibilities = 16
2B and 2G= 6/16=3/8
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 [#permalink] New post 10 Mar 2005, 07:22
Total possibilities = 2*2*2*2=16

Poss of 2 grls and 2 boys

bbgg
bgbg
bggb
ggbb
gbgb
gbbg

=6/16=3/8
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 [#permalink] New post 10 Mar 2005, 09:57
MA wrote:
it is 3/8.
2B and 2G = (4c2+4c2) = 6
total possibilities = 16
2B and 2G= 6/16=3/8


I now understand that the answer should be 3/8, however (4c2+4c2)=12, not six...

number of ways of getting two girls out of 4 babies
4c2 = 6
number of ways of getting 2 boys out of remaining 2 boys
2c2 = 1

hence we have (6*1)/(1/2^4) = 3/8

great refresher...
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 [#permalink] New post 10 Mar 2005, 17:06
HongHu wrote:
Antmavel wrote:
3/8*3/8 for me ->prob is 9/64 to get exactly 2 boys and 2 girls

what's the OA ? Hong Hu ?

Why 3/8*3/8?


:? I need some sleep....
  [#permalink] 10 Mar 2005, 17:06
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