Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

22 Jan 2010, 23:42

3

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

25% (medium)

Question Stats:

74% (01:07) correct 26% (01:24) wrong based on 218 sessions

HideShow timer Statistics

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?

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

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 \(\frac{4!}{2!2!}\).

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

Show Tags

06 Jul 2010, 11:06

4

This post received KUDOS

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

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

Show Tags

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

Answer : D

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)

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

Show Tags

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.

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

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.

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

Show Tags

28 Feb 2015, 01:32

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

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.

-- Optimus Prep's GMAT On Demand course for only $299 covers all verbal and quant. concepts in detail. Visit the following link to get your 7 days free trial account: http://www.optimus-prep.com/gmat-on-demand-course

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

Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772 Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html

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.

Answer: D
_________________

Jeffery Miller Head of GMAT Instruction

GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions