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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]

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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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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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23 Jan 2010, 01:24
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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; Equal number of boys & girls [#permalink]

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06 Jul 2010, 11:06
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Hussain15 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

So we are asked - what is probability of selecting 2 boys and 2 girls right:

We were not told about the actual selection, hence Before finding the probability, we should find what the different ways in which we can select 4 children.
So select 1 boy, then 1 boy, then 1 girl and then 1 girl or you can also select 1 B, 1 G, 1 B and 1 G.
Basically selecting 4 from BBGG. ie. 4 * 3 * 2 *1 / (2! * 2!) ways of making the selection. = 6.

Probability of selecting 1st boy = 3/6
Probability of selecting 2nd boy = 2/5 (as we have already selected one boy above)
Probability of selecting 1st Girl = 3/4
Probability of selecting 2nd Girl = 2/3

Multiply all of above with different ways of selecting the children =
P (selecting 2 boys and 2 girls) = 3/6 * 2/5 * 3/4 * 2/3 * 6 = 3/5

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Re: Probability; Equal number of boys & girls [#permalink]

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06 Jul 2010, 12:34
surjoy wrote:
Hussain15 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

So we are asked - what is probability of selecting 2 boys and 2 girls right:

We were not told about the actual selection, hence Before finding the probability, we should find what the different ways in which we can select 4 children.
So select 1 boy, then 1 boy, then 1 girl and then 1 girl or you can also select 1 B, 1 G, 1 B and 1 G.
Basically selecting 4 from BBGG. ie. 4 * 3 * 2 *1 / (2! * 2!) ways of making the selection. = 6.

Probability of selecting 1st boy = 3/6
Probability of selecting 2nd boy = 2/5 (as we have already selected one boy above)
Probability of selecting 1st Girl = 3/4
Probability of selecting 2nd Girl = 2/3

Multiply all of above with different ways of selecting the children =
P (selecting 2 boys and 2 girls) = 3/6 * 2/5 * 3/4 * 2/3 * 6 = 3/5

There is actually much simpler approach for this problem.

P (selecting 2 boys and 2 girls) = (No. of ways of selecting 2 boys out of 3 * no. of ways of selecting 2 girls out of 3) / Total ways of selecting 4 out of 6 children
= 3C2 * 3C2 / 6C4 = 3/5 (D)

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Re: Probability; Equal number of boys & girls [#permalink]

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06 Jul 2010, 12:41
Fairly straightforward question I think.

Ways to select 2 boys out of 3 = 3C2 = Ways to select 2 girls out of 3

Total ways to select 4 children = 6C4

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

Hope this helps!

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Re: Probability; Equal number of boys & girls [#permalink]

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06 Jul 2010, 13:11
any bible for the GMAT probability and combination? i read the MGMAT, it helps a little...any better books out there?

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Re: Probability; Equal number of boys & girls [#permalink]

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08 Jul 2010, 23:56
tt11234 wrote:
any bible for the GMAT probability and combination? i read the MGMAT, it helps a little...any better books out there?

Probability & Combination questions are not so common in GMAT. Hardly one can see 2 or max 3. So try to give your valuable time to the remaining 98% area of GMAT Quantative section. The concepts covered in MGMAT probability section are sufficient to answer a normal GMAT question. If you will go for a bible of GMAT probability, you will merely waste your time. So use this time to cover the topics which are most common in GMAT like number properties, Word problems & inequalities.

Best of luck.
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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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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]

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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]

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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]

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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]

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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: Probability; Equal number of boys & girls [#permalink]

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17 Jun 2015, 02:18
selecting equal no of boys and girl
selecting girl--> 3c2
selecting boy--> 3c2
total outcomes-->6c4
probab=3c2*3c2/6C4=3/5

Ans-->D
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Re: From a group of 3 boys and 3 girls, 4 children are to be [#permalink]

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17 Jun 2015, 03:58
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

ALTERNATIVELY

Probability = 1 -(Unfavorable Outcomes / Total Outcomes)

Total Outcomes = ways of selecting 4 out of 6 children = 6C4 = 15

Favorable Outcomes = 2 boys and 2 girls selected out of 3 boys and 3 girls
i.e. Unfavorable Outcomes = 3 boys and 1 girls selected out of 3 boys and 3 girls OR 1 boys and 3 girls selected out of 3 boys and 3 girls

i.e. Unfavorable Outcome_1 = 3 boys and 1 girls selected out of 3 boys and 3 girls = 3C3 * 3C1 = 1*3 = 3
and Unfavorable Outcome_2 = 1 boys and 3 girls selected out of 3 boys and 3 girls = 3C1 * 3C3 = 3*1 = 3

Probability = 1 -[(3+3) / 15] = 1 - [6/15] = 9/15 = 3/5
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Re: From a group of 3 boys and 3 girls, 4 children are to be [#permalink]

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07 Dec 2017, 09:32
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

We are given that from a group of 3 boys and 3 girls, 4 children are to be randomly selected. We need to determine the probability that an equal numbers of boys and girls will be selected, that is, the probability that two boys and two girls are selected.

We can use combinations to determine the number of favorable outcomes (that 2 boys and 2 girls are selected) and the total number of outcomes (that 4 children are selected from 6 children).

Let’s first determine the number of ways we can select 2 boys from 3 boys and 2 girls from 3 girls.

# of ways to select 2 boys from a total of 3 boys: 3C2 = 3

# of ways to select 2 girls from a total of 3 girls: 3C2 = 3

Thus, the number of ways to select 2 girls and 2 boys = 3 x 3 = 9.

Now we can determine the total number of ways to select 4 children from a total of 6 children.

6C4 = (6 x 5 x 4 x 3)/(4 x 3 x 2 x 1) = 3 x 5 = 15

Thus, the probability of selecting an equal number of girls and boys is 9/15 = 3/5.

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Re: From a group of 3 boys and 3 girls, 4 children are to be   [#permalink] 07 Dec 2017, 09:32
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