Friendship Probability

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aatech: Master | Next Rank: 500 Posts; Posts: 192; Joined: Tue Apr 29, 2008 7:41 am; Thanked: 14 times

by aatech » Tue Jun 03, 2008 1:00 pm

In a room filled with 7 people, 4 people have exactly 1 friend in the room and 3 people have exactly 2 friends in the room (Assuming that friendship is a mutual relationship, i.e. if John is Peter's friend, Peter is John's friend). If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends?
5/21
3/7
4/7
5/7
16/21

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Source: Beat The GMAT — Problem Solving | Tue Jun 03, 2008 1:00 pm

egybs: Master | Next Rank: 500 Posts; Posts: 178; Joined: Wed May 14, 2008 3:51 pmFollowed by: 1 member; Thanked: 16 times; GMAT Score:99%+

by egybs » Tue Jun 03, 2008 1:11 pm

(3/7)*(4/6) if you pick one of the popular people first (you have a 3/7 chance) he/she is then NOT friends with 4/6 of the remaining people.

(4/7)*(5/6) if you pick one of the less popular people first (you have a 4/7 chance) he/she is then NOT friends with 5/6 remaining people.

So, (3/7)*(4/6)+(4/7)*(5/6) = 32/42 = 16/21

OA is E?

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aatech: Master | Next Rank: 500 Posts; Posts: 192; Joined: Tue Apr 29, 2008 7:41 am; Thanked: 14 times

by aatech » Wed Jun 04, 2008 9:34 am

OA is E

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molt_llest: Senior | Next Rank: 100 Posts; Posts: 69; Joined: Sat May 10, 2008 3:05 pm; Thanked: 1 times

by molt_llest » Wed Jun 04, 2008 10:24 am

egybs wrote:(3/7)*(4/6) if you pick one of the popular people first (you have a 3/7 chance) he/she is then NOT friends with 4/6 of the remaining people.

(4/7)*(5/6) if you pick one of the less popular people first (you have a 4/7 chance) he/she is then NOT friends with 5/6 remaining people.

So, (3/7)*(4/6)+(4/7)*(5/6) = 32/42 = 16/21

OA is E?

I don't get the seocond statement. If tou pick one of the less popular I will have 6 people, which 3 are not friends (I think so), And you say that there is 5 people who are not friends, why?

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netigen: Legendary Member; Posts: 631; Joined: Mon Feb 18, 2008 11:57 pmFollowed by: 3 members; Thanked: 29 times

by netigen » Wed Jun 04, 2008 10:42 am

The distribution will be like -

AB - each friend of the other
CD - each friend of the other
EFG - E friend of F & G, F friend of G and E, G friend of E and F

Probability = (1 from Group 1 x 1 from G2 + 1 from Group 1 x 1 from G3 + 1 from Group 2 x 1 from G3)/ 7C2

Probability = (2C1 x 2C1 + 2C1 x 3C1 + 2C1 x 3C1)/7C2

= (4+6+6)/21 = 16/21

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