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

Question

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

Bunuel · Accepted Answer

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!} .

Answer: D.

surjoy · Answer

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

Answer : D

xcusemeplz2009 · Answer

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

DmitryFarberMPrep · Answer

Stringworm, your list of possibilities is complete--it just doesn't add up to 14 because it's the answer to a different question. You have shown all the ways to put 2 indistinguishable boys and 2 indistinguishable girls in order. What we want for our denominator is a list of all the ways we can choose 4 particular (distinguishable) children out of 3 boys and 3 girls. We aren't concerned with what order the children appear in, but we  are  concerned with which children are selected. The list looks like this:

(g1)(g2)(b1)(b2)
(g1)(g2)(b1)(b3)
(g1)(g2)(b2)(b3)
(g1)(g3)(b1)(b2)
(g1)(g3)(b1)(b3)
(g1)(g3)(b2)(b3)
(g2)(g3)(b1)(b2)
(g2)(g3)(b1)(b3)
(g2)(g3)(b2)(b3)
(b1)(all 3 girls)
(b2)(all 3 girls)
(b3)(all 3 girls)
(g1)(all 3 boys)
(g2)(all 3 boys)
(g3)(all 3 boys)

Note that the first 9 choices are the ones we want in the numerator. Also note that we would  not  want to spend time doing this on the test. If you can't find a faster way to do it, drop the problem!

EMPOWERgmatRichC · Answer

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

Final Answer:  D

GMAT assassins aren't born, they're made,
Rich

GMATinsight · Answer

ALTERNATIVELY

Probability = 1 -(Unfavorable Outcomes / Total Outcomes)

Total Outcomes = ways of selecting 4 out of 6 children = 6 C 4 = 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 = 3 C 3 * 3 C 1 = 1*3 = 3
and Unfavorable Outcome_2 = 1 boys and 3 girls selected out of 3 boys and 3 girls = 3 C 1 * 3 C 3 = 3*1 = 3

Probability = 1 -  = 1 -   = 9/15 = 3/5

JeffTargetTestPrep · Answer

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.

Answer: D

Vibhatu · Answer

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

Answer : D

Alternatively....

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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DanTheGMATMan · Answer

Combinations is definitely the way to go here. Select 2 out of 3 boys and 2 out of 3 girls and put that over the total 4 out of 6 children:
 https://www.youtube.com/watch?v=F_YI-xaumtM

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