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]
22 Jan 2010, 23:42
1
This post was BOOKMARKED
00:00
A
B
C
D
E
Difficulty:
15% (low)
Question Stats:
74% (02:12) correct
26% (01:26) wrong based on 147 sessions
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]
06 Jul 2010, 11:06
1
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]
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]
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.
Re: Probability problem [#permalink]
24 Oct 2010, 23:14
2
This post received KUDOS
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
Re: From a group of 3 boys and 3 girls, 4 children are to be [#permalink]
14 Jan 2015, 00:44
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________
Re: From a group of 3 boys and 3 girls, 4 children are to be [#permalink]
27 Feb 2015, 13:15
1
This post received KUDOS
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.
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.
Re: From a group of 3 boys and 3 girls, 4 children are to be [#permalink]
17 Jun 2015, 03:58
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
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
Classroom Centre Address: GMATinsight 107, 1st Floor, Krishna Mall, Sector-12 (Main market), Dwarka, New Delhi-110075 ______________________________________________________ Please press the if you appreciate this post !!
gmatclubot
Re: From a group of 3 boys and 3 girls, 4 children are to be
[#permalink]
17 Jun 2015, 03:58
The “3 golden nuggets” of MBA admission process With ten years of experience helping prospective students with MBA admissions and career progression, I will be writing this blog through...
You know what’s worse than getting a ding at one of your dreams schools . Yes its getting that horrid wait-listed email . This limbo is frustrating as hell . Somewhere...