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rajesh_inf · 2009 September 8th, 04:47 AM

Need explanation for this question. OA will be posted later.

In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT siblings?

A.) 5/21

B.) 3/7

C.) 4/7

D.) 5/7

E.) 16/21

kaunglay · 2009 September 9th, 08:38 AM

I choose E!

madboy · 2009 September 10th, 05:12 AM

First we need to understand the grouping of these 7 people. Since 4 people each has exactly one sibling, they can together form only 2 sibling pairs, represented diagramatically below, where A and B are a sibling pair and C and D another:
A-B
C-D
Now since of the remaining 3 people, each has exactly 2 siblings, these 3 people must be each others' siblings:
E-F-G
So now let's calculate the number of ways 2 persons can be selected such that they ARE siblings. The set of possible outcomes are : A-B, C-D, E-F, F-G and G-E. So we can do this in total 5 ways
Total number of ways in which any 2 people can be chosen from the group = 7C2 = 7!/(2!5!) = 21 (If you do not understand this part you need to review Combination theory)
Therefore, probability that 2 persons selected at random will be siblings = 5/21
And hence the probability that they ARE NOT siblings = complement of the above = 1 - 5/21 = 16/21 (E)

sh_vivek · 2009 September 11th, 06:47 AM

Group I : a,b,c
Group II: d,e
Group III: f,g

so required probablity =
(one person from I AND one person from II) OR
(one person from II AND one person from I) OR
(one person from I AND one person from III) OR
(one person from III AND one person from I) OR
(one person from II AND one person from III) OR
(one person from III AND one person from II)

= (3/7*2/6)+(2/7*3/6)+(3/7*2/6)+(2/7*3/6)+(2/7*2/6)+(2/7*2/6)
= 16/21.

OR

= 1-probablity that both selected are siblings
= 1-(3C2+2C2+2C2)
=16/21

OR

number of ways of not selecting sibling = 3C1*2C1+3C1*2C1+2C1*2C1 = 16
total ways of selecting two people = 7C2 = 21
So required probabilty = 16/21

efet · 2009 September 11th, 11:07 PM

Very irritating question. I saw this question the day I started studying for gmat. I drew the correct drawing but couldnt figure the answer out. I see it now and I still cant figure the final answer :S

Madboy's explanation is correct.

2009 September 8th, 04:47 AM	#1 (permalink)
rajesh_inf Eager! Join Date: May 2008 Posts: 42	Picking Friends - Probability Need explanation for this question. OA will be posted later. In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT siblings? A.) 5/21 B.) 3/7 C.) 4/7 D.) 5/7 E.) 16/21

2009 September 10th, 05:12 AM	#3 (permalink)
madboy 650 and aiming higher Join Date: Sep 2009 Location: Lowell,MA,USA Posts: 349	First we need to understand the grouping of these 7 people. Since 4 people each has exactly one sibling, they can together form only 2 sibling pairs, represented diagramatically below, where A and B are a sibling pair and C and D another: A-B C-D Now since of the remaining 3 people, each has exactly 2 siblings, these 3 people must be each others' siblings: E-F-G So now let's calculate the number of ways 2 persons can be selected such that they ARE siblings. The set of possible outcomes are : A-B, C-D, E-F, F-G and G-E. So we can do this in total 5 ways Total number of ways in which any 2 people can be chosen from the group = 7C2 = 7!/(2!5!) = 21 (If you do not understand this part you need to review Combination theory) Therefore, probability that 2 persons selected at random will be siblings = 5/21 And hence the probability that they ARE NOT siblings = complement of the above = 1 - 5/21 = 16/21 (E) Last edited by madboy : 2009 September 11th at 01:44 AM.

2009 September 11th, 06:47 AM	#4 (permalink)
sh_vivek first attempt:timed out Join Date: Jul 2008 Posts: 842	Group I : a,b,c Group II: d,e Group III: f,g so required probablity = (one person from I AND one person from II) OR (one person from II AND one person from I) OR (one person from I AND one person from III) OR (one person from III AND one person from I) OR (one person from II AND one person from III) OR (one person from III AND one person from II) = (3/72/6)+(2/73/6)+(3/72/6)+(2/73/6)+(2/72/6)+(2/72/6) = 16/21. OR = 1-probablity that both selected are siblings = 1-(3C2+2C2+2C2) =16/21 OR number of ways of not selecting sibling = 3C12C1+3C12C1+2C12C1 = 16 total ways of selecting two people = 7C2 = 21 So required probabilty = 16/21 _ _ _ _ SIG _ _ _ _ When you want something, all the universe conspires in helping you to achieve it. --The Alchemist* Last edited by sh_vivek : 2009 September 11th at 08:17 AM.

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2009 September 9th, 08:38 AM	#2 (permalink)
kaunglay I JUST got here. Join Date: Sep 2009 Posts: 1	I choose E!

2009 September 11th, 11:07 PM	#5 (permalink)
efet This user's posts are moderated. Join Date: Aug 2008 Posts: 113	Very irritating question. I saw this question the day I started studying for gmat. I drew the correct drawing but couldnt figure the answer out. I see it now and I still cant figure the final answer :S Madboy's explanation is correct.

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