|
Author |
Message |
|
TAGS:
|
|
|
Director
Joined: 07 Nov 2004
Posts: 511
Followers: 2
Kudos [?]:
1
[0], given: 0
|
A couple want to have four babies, for each baby, 50% are [#permalink]
 16 Feb 2005, 00:48
Question Stats:
0% (00:00) correct
 0% (00:00) wrong based on 0 sessions
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?
|
|
|
|
|
|
|
Manager
Joined: 13 Feb 2005
Posts: 63
Location: Lahore, Pakistan
Followers: 1
Kudos [?]:
0
[0], given: 0
|
I think your question is incomplete.....please specify the number of babies
|
|
|
|
|
|
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5134
Location: Singapore
Followers: 9
Kudos [?]:
87
[0], given: 0
|
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
|
|
|
|
|
|
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5134
Location: Singapore
Followers: 9
Kudos [?]:
87
[0], given: 0
|
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) 
|
|
|
|
|
|
VP
Joined: 13 Jun 2004
Posts: 1135
Location: London, UK
Schools: Tuck'08
Followers: 5
Kudos [?]:
14
[0], given: 0
|
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) My verbal skills are terrible, i am really a poor french guy 
That's why I only hunt in the Quant forum 
|
|
|
|
|
|
Intern
Joined: 19 Feb 2005
Posts: 10
Followers: 0
Kudos [?]:
0
[0], given: 0
|
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 ?
|
|
|
|
|
|
Director
Joined: 13 Nov 2003
Posts: 811
Location: BULGARIA
Followers: 1
Kudos [?]:
6
[0], given: 0
|
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
|
|
|
|
|
|
SVP
Joined: 03 Jan 2005
Posts: 2322
Followers: 9
Kudos [?]:
157
[0], given: 0
|
Antmavel wrote: My verbal skills are terrible, i am really a poor french guy That's why I only hunt in the Quant forum Huh? If this is the case you should really hunt the Verbal forum more Antmavel. 
|
|
|
|
|
|
SVP
Joined: 03 Jan 2005
Posts: 2322
Followers: 9
Kudos [?]:
157
[0], given: 0
|
Please read this thread and see if it gives you some light. 
http://www.gmatclub.com/phpbb/viewtopic.php?t=14548
|
|
|
|
|
|
SVP
Joined: 03 Jan 2005
Posts: 2322
Followers: 9
Kudos [?]:
157
[0], given: 0
|
Read this and see if you could do it again, vprabhala.
http://www.gmatclub.com/phpbb/viewtopic ... 9947#89947
|
|
|
|
|
|
Director
Joined: 21 Sep 2004
Posts: 738
Followers: 1
Kudos [?]:
3
[0], given: 0
|
but it doesn't say atleast. it says.. what is the probability of getting 2 boys and 2 girls right Honghu?
|
|
|
|
|
|
Director
Joined: 21 Sep 2004
Posts: 738
Followers: 1
Kudos [?]:
3
[0], given: 0
|
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
|
|
|
|
|
|
Intern
Joined: 27 Jun 2004
Posts: 35
Followers: 0
Kudos [?]:
0
[0], given: 0
|
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.
|
|
|
|
|
|
Director
Joined: 21 Sep 2004
Posts: 738
Followers: 1
Kudos [?]:
3
[0], given: 0
|
4C2*(1/2)^2*(1/2)^2=3/8
I still don't get hte difference between exactly, atleast etc..
can Honghu please explain..
|
|
|
|
|
|
SVP
Joined: 30 Sep 2004
Posts: 1548
Location: Germany
Followers: 4
Kudos [?]:
15
[0], given: 0
|
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
|
|
|
|
|
|
VP
Joined: 13 Jun 2004
Posts: 1135
Location: London, UK
Schools: Tuck'08
Followers: 5
Kudos [?]:
14
[0], given: 0
|
3/8*3/8 for me ->prob is 9/64 to get exactly 2 boys and 2 girls
what's the OA ? Hong Hu ?
|
|
|
|
|
|
SVP
Joined: 25 Nov 2004
Posts: 1582
Followers: 4
Kudos [?]:
13
[0], given: 0
|
it is 3/8.
2B and 2G = (4c2+4c2) = 6
total possibilities = 16
2B and 2G= 6/16=3/8
|
|
|
|
|
|
Manager
Joined: 01 Jan 2005
Posts: 169
Location: NJ
Followers: 1
Kudos [?]:
0
[0], given: 0
|
Total possibilities = 2*2*2*2=16
Poss of 2 grls and 2 boys
bbgg
bgbg
bggb
ggbb
gbgb
gbbg
=6/16=3/8
|
|
|
|
|
|
Senior Manager
Joined: 07 Oct 2003
Posts: 380
Location: Manhattan
Followers: 1
Kudos [?]:
2
[0], given: 0
|
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...
|
|
|
|
|
|
VP
Joined: 13 Jun 2004
Posts: 1135
Location: London, UK
Schools: Tuck'08
Followers: 5
Kudos [?]:
14
[0], given: 0
|
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....
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
close
