Math questions from any Manhattan Prep GMAT Computer Adaptive Test.
james.jt.wu
 
Posts: 27
Joined: Tue May 06, 2008 10:47 pm
 

Re: Siblings Problem - Combinatorics

by james.jt.wu Sat Oct 09, 2010 2:46 am

Wanted to bump this question up mainly because this question is great for back-up method (thanks ron).

Method 3 to doing this problem - just list them!

First, total number of ways of picking 2 people out of 7 = (7*6)/2! = 21.

Next, the number of ways to pick a pair of sibling:

Use a letter to denote each person in the group:

A-B = first 2-sibling group
C-D= second 2-sibling group
E-F-G = 3-sibling group

So your pairs are:

AB
CD
EF
FG
EG

(remember here the positions are interchangeable... so AB and BA are the same pair)

5 total ways to pick two people who are siblings, so you probability of picking a sibling is 5/21.

Applying the 1-x rule: 1 - (5/21) = 16/21 probability of NOT sibling.

Bravo MGMAT - very nice question.
mschwrtz
ManhattanGMAT Staff
 
Posts: 498
Joined: Tue Dec 14, 2004 1:03 pm
 

Re: Siblings Problem - Combinatorics

by mschwrtz Sun Oct 17, 2010 11:53 am

Glad you liked it James. Maybe we should add Ron's note to the CAT explanation.
WarrenGMAT
Students
 
Posts: 1
Joined: Sat Nov 06, 2010 5:04 pm
 

Re: Siblings Problem - Combinatorics

by WarrenGMAT Tue Nov 23, 2010 4:51 pm

4/7 you pick someone with 1 sibling. 3/7 you pick someone with 2 siblings.

If you pick a 1 sibling person, he has one sibling in the deck of 6 remaining people: 4/7 x 1/6 = 4/42 = 2/21

If you pick a 2 sibling person: 3/7 x 2/6 = 6/42 = 3/21

Add them: 5/21 says you get a sibling pair.

1 - 5/21 = 16/21 says you don't.
jnelson0612
ManhattanGMAT Staff
 
Posts: 2664
Joined: Fri Feb 05, 2010 10:57 am
 

Re: Siblings Problem - Combinatorics

by jnelson0612 Tue Nov 23, 2010 6:56 pm

Nice work everyone! Thank you to our contributors.
Jamie Nelson
ManhattanGMAT Instructor
gmatwork
Course Students
 
Posts: 185
Joined: Wed Dec 31, 1969 8:00 pm
 

Re: Siblings Problem - Combinatorics

by gmatwork Fri Dec 02, 2011 2:40 pm

Can't we solve this problem as -

P(no siblings selected) = 1 - P( siblings selected)

P( siblings selected) = P(siblings from group 1) + P (Siblings from group 2) + P (siblings from group 3)

groups - are the three groups based on having two and three sibling pairs as said above in the posts

= 2C2/7C2 + 2C2/7C2 + 3C2/7C2
= 5/21

Then P(no siblings) = 1- 5/21 = 16/21

Please point out if I missed anything?
Last edited by gmatwork on Fri Dec 02, 2011 2:45 pm, edited 1 time in total.
gmatwork
Course Students
 
Posts: 185
Joined: Wed Dec 31, 1969 8:00 pm
 

Re: Siblings Problem - Combinatorics

by gmatwork Fri Dec 02, 2011 2:41 pm

Can't we solve this problem as -

P(no siblings selected) = 1 - P( siblings selected)

P( siblings selected) = P(siblings from group 1) + P (Siblings from group 2) + P (siblings from group 3)

groups - are the three groups based on having two and three sibling pairs as said above in the posts

= 2C2/7C2 + 2C2/7C2 + 3C2/7C2
= 5/21

Then P(no siblings) = 1- 5/21 = 16/21

Please point out if I missed anything?
tim
Course Students
 
Posts: 5665
Joined: Tue Sep 11, 2007 9:08 am
Location: Southwest Airlines, seat 21C
 

Re: Siblings Problem - Combinatorics

by tim Tue Dec 13, 2011 2:48 am

Totally appropriate to do it this way, and you did it perfectly!
Tim Sanders
Manhattan GMAT Instructor

Follow this link for some important tips to get the most out of your forum experience:
https://www.manhattanprep.com/gmat/forums/a-few-tips-t31405.html