BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

probability

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Senior | Next Rank: 100 Posts
Posts: 46
Joined: Fri Aug 19, 2011 12:13 am
Location: Delhi, India
Followed by:1 members

probability

by romitvsingh » Mon Nov 28, 2011 8:08 pm
A couple decides to have 4 children. If they suceed in having 4 children & each is equally likely to be a boy or a girl, what is the probaility that the will have exactly 2 girls or 2 boys?

A) 3/8
B) 1/4
C) 3/16
D) 1/8
E) 1/16

Pls explain in detail
Eat, sleep GMAT
Top
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pm
Location: Milpitas, CA
Thanked: 1854 times
Followed by:523 members
GMAT Score:770

by Anurag@Gurome » Mon Nov 28, 2011 8:38 pm
romitvsingh wrote:A couple decides to have 4 children. If they suceed in having 4 children & each is equally likely to be a boy or a girl, what is the probaility that the will have exactly 2 girls or 2 boys?

A) 3/8
B) 1/4
C) 3/16
D) 1/8
E) 1/16

Pls explain in detail
Possibilities of having 2 boys and 2 girls = BBGG, GGBB, BGBG, GBGB, BGGB, GBBG, which means 6 possibilities.

Total no. of ways in which 2 children can be born = 2^4 = 16
So, required probability = 6/16 = 3/8

The correct answer is A.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 461
Joined: Tue May 10, 2011 9:09 am
Location: pune
Thanked: 36 times
Followed by:3 members

by amit2k9 » Mon Nov 28, 2011 11:06 pm
(1/2)^4 * 4C2 = 3/8
4 kids with probability of 50 for a girl of a boy.
4C2 for exactly two boys or two girls.
For Understanding Sustainability,Green Businesses and Social Entrepreneurship visit -https://aamthoughts.blocked/
(Featured Best Green Site Worldwide-https://bloggers.com/green/popular/page2)
Top
User avatar
Senior | Next Rank: 100 Posts
Posts: 35
Joined: Sun Dec 02, 2007 1:06 pm

by jaguar123 » Sat Dec 03, 2011 12:22 am
One doubt.. what is the mistake with this method.

Total number of combinations - without oder -5.

1. 4B 0G - 0B,4G
2. 3B 1G - 1B 3G
3. 2B,2G

there is only one combination that works fine - 2B,2G - 1/5. Please help.
Top
Legendary Member
Posts: 1084
Joined: Fri Apr 15, 2011 2:33 pm
Thanked: 158 times
Followed by:21 members

by pemdas » Sat Dec 03, 2011 1:15 am
this q. can be simplified if we ask "When tossing a coin 4 times, how many times we have 2 tails and 2 heads exactly?" Try solving this way and it will be much easier.

Now with your doubt again, if you consider coin instead of child. Let me ask you how you can know if the coin tossed by you will land in 4 tails (or heads) and 0 heads (or tails), 3 tails (or heads) and 3 heads (or tails), ...?
You've prearranged the order of not "the only possible ways" but the order of "samples: 4B 0G, 4G 0B, 3B 1G, 3G 1B ... etc,"
Each sample set may have different probability and will be the baseline for a new question :)

stick with coins and try to think further about this question.
jaguar123 wrote:One doubt.. what is the mistake with this method.

Total number of combinations - without oder -5.

1. 4B 0G - 0B,4G
2. 3B 1G - 1B 3G
3. 2B,2G

there is only one combination that works fine - 2B,2G - 1/5. Please help.
Success doesn't come overnight!
Top
Post new topic Post Reply
• Page 1 of 1