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

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From a group of 3 boys and 3 girls, 4 children are to be [#permalink]  22 Jan 2010, 23:42
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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
[Reveal] Spoiler: OA
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Re: probability [#permalink]  23 Jan 2010, 00:58
prob reqd= 1- probab of not having eq numbers

prob of not having eq no's is = prob of having 3 Boys and 1 girl or prob of having 3 girls and 1 boy
=(1* 3/15) +(3/15 * 1)=2/5

reqd prob= 1-2/5=3/5
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Re: probability [#permalink]  23 Jan 2010, 01:24
Expert's post
2
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xcusemeplz2009 wrote:
prob reqd= 1- probab of not having eq numbers

prob of not having eq no's is = prob of having 3 Boys and 1 girl or prob of having 3 girls and 1 boy
=(1* 3/15) +(3/15 * 1)=2/5

reqd prob= 1-2/5=3/5

In this case it's easier to calculate directly what is asked. Equal # of boys and girls out of 4 means 2 boys and 2 girls.

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

$$\frac{3C2*3C2}{6C4}=\frac{3}{5}$$

OR: $$\frac{4!}{2!2!}*\frac{3}{6}*\frac{2}{5}*\frac{3}{4}*\frac{2}{3}=\frac{3}{5}$$.

Here we are counting the probability of BBGG, this combination can occur in different # of ways: BGBG, GGBB, ... Total # of ways would be the # permutations of the letters BBGG, which is $$\frac{4!}{2!2!}$$.

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Re: Probability problem [#permalink]  24 Oct 2010, 23:14
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Merging similar topics ....

monirjewel wrote:
From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the probability that equal number of boys and girls will be selected?

Total ways to pick children = C(6,4) = 15

Ways to pick 2 boys & 2 girls = C(3,2) x C(3,2) = 9

Probability = 9/15 = 3/5
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Re: probability.... [#permalink]  30 Nov 2011, 00:34
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There is only 1 way of selecting equal number of boys and girls i.e. 2 boys and 2 girls.

Boys- 3C2 Girls- 3C2
Total possible selections: 6C4

Probability= (3C2*3C2)/6C4 = 3/5
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Re: From a group of 3 boys and 3 girls, 4 children are to be [#permalink]  14 Jan 2015, 00:44
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Re: From a group of 3 boys and 3 girls, 4 children are to be [#permalink]  14 Feb 2015, 01:11
The only way not presentsed is reverse combination, which is:

denominator: 6C4=15

numerator: (3C3*1C3)+(3C3*1C3)=6

so 1 - 6/15=9/15=3/5

D
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Re: From a group of 3 boys and 3 girls, 4 children are to be [#permalink]  14 Feb 2015, 05:31
Can somebody explain why we need to use combinatorics here and not solely probability (1/6 * 4) ?
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Re: From a group of 3 boys and 3 girls, 4 children are to be [#permalink]  27 Feb 2015, 08:53
1st step: randomly select 4 from 6 = C(6 4) = 6! / 4!2! = 15
2nd step: select 2 girls from 3 girls = C(3 2) = 3! / 2!1! = 3
3rd step: select 2 boys from 3 boys = C(3 2) = 3! / 2!1! = 3
(3+3) / 15 = 3/5
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Re: From a group of 3 boys and 3 girls, 4 children are to be [#permalink]  27 Feb 2015, 13:15
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Expert's post
Hi LaxAvenger,

You COULD calculate the answer to this question using 'probability math', but the calculation would be far MORE complicated than what you wrote down. Here's how it would work....

First, you have to account for all of the different 'ways' to get 2 boys and 2 girls. Assuming that the children are chosen one at a time, here are the options that "fit" what we're looking for:

BBGG
BGBG
BGGB
GBBG
GBGB
GGBB

Using the first example, here is the probability of THAT EXACT sequence occurring:
BBGG = (3/6)(2/5)(3/4)(2/3) = 36/360 = 1/10

Each of the other 5 options will yield the exact SAME probability....
eg
BGBG = (3/6)(3/5)(2/4)(2/3) = 36/360 = 1/10

So we have 6 different options that each produce a 1/10 chance of occurring.

6(1/10) = 6/10 = 3/5

[Reveal] Spoiler:
D

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Re: From a group of 3 boys and 3 girls, 4 children are to be [#permalink]  28 Feb 2015, 01:32
Expert's post
vaivish1723 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

Total children = 6
We need 2 boys and 2 girls.
Total ways of selecting boys = 3C2 = 3
Total ways of selecting girls = 3C2 = 3
Total ways of selecting 4 children = 6C4 = 15

Required probability = (3 * 3)/15
= 3/5
Hence option D.

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Re: From a group of 3 boys and 3 girls, 4 children are to be   [#permalink] 28 Feb 2015, 01:32
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