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From a group of 3 boys and 3 girls, 4 children are to be

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Director
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From a group of 3 boys and 3 girls, 4 children are to be [#permalink] New post 29 Apr 2006, 20:46
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From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the probability that equal numbers of boys and girls will be selected?

A. 1/10
B. 4/9
C. 1/2
D. 3/5
E. 2/3
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Re: PS: Prob (Boys/Girls) [#permalink] New post 29 Apr 2006, 22:23
Professor wrote:
M8 wrote:
From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the probability that equal numbers of boys and girls will be selected?

A. 1/10
B. 4/9
C. 1/2
D. 3/5
E. 2/3


= 2x3c2/6c4


Prof what about this one 3C2x3C2/6C4? :wink:
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Re: PS: Prob (Boys/Girls) [#permalink] New post 30 Apr 2006, 00:41
TeHCM wrote:
Agree with 3C2x3C2/6C4

So 3/10?


Be careful. :wink:
Approach is correct but the answer is not.
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 [#permalink] New post 30 Apr 2006, 01:26
as order does not matter we can simply use

Probability of 1 st girl * Probability of 2 gir(provided 1st was girl) * Probability of 1st boy * Probability of 2nd boy

3/6*2/5*3/4*2/3 =>3*2*3*2/6*5*4*3 = 1/10
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Re: PS: Prob (Boys/Girls) [#permalink] New post 30 Apr 2006, 05:48
M8 wrote:
TeHCM wrote:
Agree with 3C2x3C2/6C4

So 3/10?


Be careful. :wink:
Approach is correct but the answer is not.


hmmmmmmmmm......................... i added instead of multiplying.

agree with this one
=(3c2 x 3c2)/6c4=3/5
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 [#permalink] New post 30 Apr 2006, 08:47
find all the combinations of 2 boys and two girls:
ways we can choose 2 of 3 = 2C3 = 3. 3*3 = 9

total combinations of 4C6 = 15.

9/15 = 3/5
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Re: PS: Prob (Boys/Girls) [#permalink] New post 30 Apr 2006, 19:14
trivikram wrote:
M8 wrote:
Professor wrote:
M8 wrote:
From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the probability that equal numbers of boys and girls will be selected?

A. 1/10
B. 4/9
C. 1/2
D. 3/5
E. 2/3

= 2x3c2/6c4

Prof what about this one 3C2x3C2/6C4? :wink:

But why should we multiply here?

3 boys = a, b and c
3 girls = x, y and z

no of ways 2 boys can be chosen = ab, bc, and ca
no of ways 2 girls can be chosen = xy, yz, and zx

no of ways 2boys and 2 girls can be chosen = (ab+xy), (ab+yz), ........................ and (ca+zx) = 9 ways

no of ways 4 out of 6 can be chosen = 6c4 = 15

so the prob = 9/15 = 3/5
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 [#permalink] New post 02 May 2006, 00:23
Only way is to pick 2 girls and 2 boys.

# of ways to pick 2 girls = 3C2 = 3
# of ways to pick 2 boys = 3C2 = 3
# of ways to pick 2 girls and 2 boys = 3*3 = 9
# of ways to pick 4 children = 6C4 = 15

P = 9/15 = 3/5
  [#permalink] 02 May 2006, 00:23
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