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

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

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)

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!

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.

Merging similar topics ....



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

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!

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

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

Can somebody explain why we need to use combinatorics here and not solely probability (1/6 * 4) ?

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
Answer is D

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:

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

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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selecting equal no of boys and girl
selecting girl--> 3c2
selecting boy--> 3c2
total outcomes-->6c4
probab=3c2*3c2/6C4=3/5

Ans-->D

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

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

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?

Probability = Favourable outcomes/Total no. of outcomes = [3C2x3C2]/6C4 = 3x3/15 = 3/5

Hence D