Problem 4 - Boys and Girls
What is the probability that a man having four children has two girls and two boys, given that he has at least one girl.
The answer choices are a) 0.375 b) 0.5 c) 0.4 d) None of these
Answers by tomorrow morning
The answer choices are a) 0.375 b) 0.5 c) 0.4 d) None of these
Answers by tomorrow morning
Gyan-ee 11:07 PM
5 Comments:
hi,
i am a confused with two ways of going about this , please help
1)
assuming the first kid is a girl , then probab of the second being a girl and other two guys is 1/2*1/2*1/2
similarly for the third being a girl and second and the last guy
1/2*1/2*1/2
and for the fourth again .125
so in all the probab = .375 ( ANS)
2)
since that guy has atleast one girl
the possibilites are
GBBB
GGBB
GGGB
GGGG
BBBB
so ANS = 1 /4
now i wonder which is wrong :(
-- alohabrij
The answer is .4
This was a very simple problem. The key in probability is to keep your thinking as simple as the problem allows one to.
There are 16 permutations for the four children.
2*2*2*2 = 16.
GGGG
BBBB
BBGG
BGBG
BGGB
GGBB
GBGB
GBBG
GBBB
BGBB
BBGB
BBBG
GGGB
GGBG
GBGG
BGGG
Now, since it is already given that one of them is a girl, the permutation of BBBB is ruled out. That leaves 15 permutations.
In that count the ones that have just one girl. That would be six.
BBGG, BGBG, BGGB, GGBB, GBGB, GBBG
So the probability is 6/15 = 0.4
Alohabrij,
The mistake in your first approach was that you assumed the "first" kid is a girl
The second approach was right, but the permutations were not complete.
This comment has been removed by the author.
Total cases in case nothing is known abt children ==16
but we know that one kid is a girl == total cases ==15
and possibilities == (GGGB) (GGBB) (BBBG)
now (GGGB),(GGBG),(GBGG) or (BGGG) ==4 cases x2 for (BBBG)==8
==> left cases =7-1 BBBB==6
6/15=2/5=.4
Right again.
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