• Magoosh
Study with Magoosh GMAT prep

Available with Beat the GMAT members only code

• Free Trial & Practice Exam
BEAT THE GMAT EXCLUSIVE

Available with Beat the GMAT members only code

• Reach higher with Artificial Intelligence. Guaranteed
Now free for 30 days

Available with Beat the GMAT members only code

• Get 300+ Practice Questions

Available with Beat the GMAT members only code

• Award-winning private GMAT tutoring
Register now and save up to \$200

Available with Beat the GMAT members only code

• Free Veritas GMAT Class
Experience Lesson 1 Live Free

Available with Beat the GMAT members only code

• 5 Day FREE Trial
Study Smarter, Not Harder

Available with Beat the GMAT members only code

• Free Practice Test & Review
How would you score if you took the GMAT

Available with Beat the GMAT members only code

• 1 Hour Free
BEAT THE GMAT EXCLUSIVE

Available with Beat the GMAT members only code

• 5-Day Free Trial
5-day free, full-access trial TTP Quant

Available with Beat the GMAT members only code

Difficult Math Problem #95 - Probability

This topic has 3 member replies
800guy Master | Next Rank: 500 Posts
Joined
27 Jun 2006
Posted:
354 messages
Followed by:
5 members
11

Difficult Math Problem #95 - Probability

Fri Feb 09, 2007 10:32 am
from diff math problems doc. oa coming when some people answer with explanations..

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

Top Member

Roland2rule Legendary Member
Joined
30 Aug 2017
Posted:
713 messages
Followed by:
4 members
Sun Nov 05, 2017 11:09 am
probability of event n happening is given as
$$\left(\Pr\left(n\right)=\ \frac{\left(no.\ of\ required\ outcomes\right)}{\left(no.\ of\ possible\ outcomes\right)}\right)$$
In a group of 3 girls and 3 boys, equal number of boys and girls can be selected if 2 boys AND 2 girls are selected.
let x and y be the probability of selecting 2 boys and 3 boys and the probability of selecting 2 girls from 3 girls, respectively.
$$\Pr\left(x\right)=\ \frac{\left(no.\ of\ boys\ required\right)}{\left(total\ no.\ of\ boys\right)}=\frac{2}{3}$$
The same way;
$$\Pr\left(y\right)=\ \frac{\left(no.\ of\ girls\ required\right)}{\left(total\ no.\ of\ girls\right)}=\frac{2}{3}$$
therefore,
$$\Pr\left(x\ AND\ y\right)=\frac{2}{3}\cdot\frac{2}{3}=\ \frac{4}{9}$$

800guy Master | Next Rank: 500 Posts
Joined
27 Jun 2006
Posted:
354 messages
Followed by:
5 members
11
Mon Feb 12, 2007 3:25 pm
OA:

Total number of ways of selecting 4 children = 6C4 = 15

with equal boys and girls. => 2 boys and 2 girls. => 3C2 * 3C2 = 9.

Hence p = 9/15 = 3/5

banona Senior | Next Rank: 100 Posts
Joined
30 Dec 2006
Posted:
38 messages
1
Fri Feb 09, 2007 12:30 pm
I think answer is (3/5). Why ?

We have a group of 3 boys and 3 girls, 4 children are to be randomly selected.
all possible outcomes are of number of combinaison of sets of 4 members from a set of six numbers = 6!/2!*4! = 15

The assess the probability that an event of equal numbers of boys and girls will be selected, we have to count the number of possible sets that countain 2 boys and two girls;
there are many ways to do such computation, I would go for the following one using combinaisons and multiplication. Rephrase : We have a number of event sets equal to [( number of sets of two girls from 3)*( number of sets of two boys from 3)] = 3*3 = 9

Therfore probability is : 9/15 = 3/5

I hope there I was clear;
looking forward to reading some feed-backs

Banona

A. 1/10

B. 4/9

C. 1/2

D. 3/5

E. 2/3

Best Conversation Starters

1 lheiannie07 80 topics
2 LUANDATO 59 topics
3 ardz24 52 topics
4 AAPL 45 topics
5 Roland2rule 43 topics
See More Top Beat The GMAT Members...

Most Active Experts

1 Rich.C@EMPOWERgma...

EMPOWERgmat

133 posts
2 Brent@GMATPrepNow

GMAT Prep Now Teacher

131 posts
3 GMATGuruNY

The Princeton Review Teacher

130 posts
4 Scott@TargetTestPrep

Target Test Prep

118 posts
5 Jeff@TargetTestPrep

Target Test Prep

114 posts
See More Top Beat The GMAT Experts