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KK
 
 

probability of 2 girls out of 1/3

by KK Fri Jul 04, 2008 7:41 pm

While working on a SG problem, i thought of this question that troubles me in probability. If 1/3 of the room is filled with girls, rest being boys, if 2 girls are selected randomly what is the probability that they are both girls.

Also i would like to know if i need to refer any other resources apart from SG and OG for practice of probability questions, as in comparison with other topics, i found very less questions on this topic. Is there a reason why. Are they not so frequent questions on GMAT

Thanks
Guest
 
 

Re: probability of 2 girls out of 1/3

by Guest Fri Jul 11, 2008 7:17 pm

KK Wrote:While working on a SG problem, i thought of this question that troubles me in probability. If 1/3 of the room is filled with girls, rest being boys, if 2 girls are selected randomly what is the probability that they are both girls.

Also i would like to know if i need to refer any other resources apart from SG and OG for practice of probability questions, as in comparison with other topics, i found very less questions on this topic. Is there a reason why. Are they not so frequent questions on GMAT

Thanks


I think that you stated your question incorrectly. If you choose two girls, the probability that they'll both be girls is 100% unless there are come cross dressers or other things I can't mention here.


You said that 1/3rd of the room is filled with girls. This means that you have 1 girl for every 2 boys. If two people are selected, what is the probability that both are girls?

Off the bat, this question cannot be answered because if you have 1 girl and 2 boys, then the probability of having 2 girls if 0. If you have 2 girls and 6 boys, then you can actually have a chance of picking two girls. Assuming that the # of girls is greater then 1...



Ex: A
2 G, 6 B (Mary, Sue-----John, Peter, Paul, Rob, Mike, Bill)

8C2 = 28 total non repeating combinations of groups of two people

There is only one combination of 2 girls so the answer is: 1/28

Ex: B
3 G, 9 B (Mary, Sue, Lafonda-----John, Peter, Paul, Rob, Mike, Bill, Rick, Evert, Lupe)

9C2=36 total non repeating combinations of groups of two people

Mary, Sue
Mary, Lafonda
Sue Lafonda

(3 combinations of two girls)

3/36 = 1/12 combinations of two girls



This shows us that without a concrete number, you cannot answer this problem!
rfernandez
Course Students
 
Posts: 381
Joined: Fri Apr 07, 2006 8:25 am
 

by rfernandez Fri Jul 18, 2008 12:47 pm

Nice job, Guest. And some creative names, too!