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# A recent married couple decided to have four kids. What is

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VP
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A recent married couple decided to have four kids. What is [#permalink]

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25 Aug 2007, 08:40
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A recent married couple decided to have four kids. What is the probability that they will have exactly 2 boys and 2 girls?
VP
Joined: 08 Jun 2005
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25 Aug 2007, 09:04
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the probability for G or B is 1/2

1/2*1/2*1/2*1/2 = 1/16

BBGG = 1/16
BGBG = 1/16
BGGB = 1/16
GBGB = 1/16
GBBG = 1/16
GGBB = 1/16

1/16+1/16+1/16+1/16+1/16+1/16 = 6/16 = 3/8

VP
Joined: 10 Jun 2007
Posts: 1444
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Kudos [?]: 285 [0], given: 0

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25 Aug 2007, 09:29
KillerSquirrel wrote:
the probability for G or B is 1/2

1/2*1/2*1/2*1/2 = 1/16

BBGG = 1/16
BGBG = 1/16
BGGB = 1/16
GBGB = 1/16
GBBG = 1/16
GGBB = 1/16

1/16+1/16+1/16+1/16+1/16+1/16 = 6/16 = 3/8

I'm trying to see what's wrong with my method...if you can help.
I got the same answer using the method above. However, when I use this method, I get different answer:
Say if you have the first 2 child, you have have either BB,BG,GB,GG. This means that the first 2 child doesn't affect the probability. So, wouldn't the last two child pick would be 1/2 * 1/2 = 1/4?
VP
Joined: 08 Jun 2005
Posts: 1145
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25 Aug 2007, 10:18
bkk145 wrote:
KillerSquirrel wrote:
the probability for G or B is 1/2

1/2*1/2*1/2*1/2 = 1/16

BBGG = 1/16
BGBG = 1/16
BGGB = 1/16
GBGB = 1/16
GBBG = 1/16
GGBB = 1/16

1/16+1/16+1/16+1/16+1/16+1/16 = 6/16 = 3/8

I'm trying to see what's wrong with my method...if you can help.
I got the same answer using the method above. However, when I use this method, I get different answer:
Say if you have the first 2 child, you have have either BB,BG,GB,GG. This means that the first 2 child doesn't affect the probability. So, wouldn't the last two child pick would be 1/2 * 1/2 = 1/4?

I don't think you can solve this problem two kids at a time.

Director
Joined: 03 May 2007
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Schools: University of Chicago, Wharton School
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25 Aug 2007, 16:42
bkk145 wrote:
A recent married couple decided to have four kids. What is the probability that they will have exactly 2 boys and 2 girls?

2 b and 2 g = 4!/2!2! = 6
total = 2^4 = 16
prob = 6/16 = 3/8
VP
Joined: 08 Jun 2005
Posts: 1145
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Kudos [?]: 213 [0], given: 0

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25 Aug 2007, 21:51
There is a faster way to solve those kind of questions, but I think this is out of scope.

Applying the binomial distribution formula:

C(n,k)*p^k*(1-p)^(n-k)

p = the probability for a certain event = 1/2 (getting a boy)

1-P = the completing event = 1/2 (getting a girl)

n = total items in the group = 4

k = desired outcome = two boys = 2

solving

C(4,2)*(1/2)^(2)*(1/2)*(4-2) = 6*1/4*1/4 = 6/16 = 3/8

Director
Joined: 03 May 2007
Posts: 876
Schools: University of Chicago, Wharton School
Followers: 6

Kudos [?]: 208 [0], given: 7

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25 Aug 2007, 23:33
KillerSquirrel wrote:
There is a faster way to solve those kind of questions, but I think this is out of scope.

Applying the binomial distribution formula:

C(n,k)*p^k*(1-p)^(n-k)

p = the probability for a certain event = 1/2 (getting a boy)

1-P = the completing event = 1/2 (getting a girl)

n = total items in the group = 4

k = desired outcome = two boys = 2

solving

C(4,2)*(1/2)^(2)*(1/2)*(4-2) = 6*1/4*1/4 = 6/16 = 3/8

yup i was looking for the binomial formula.
25 Aug 2007, 23:33
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