Probability-siblings

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lilu: Master | Next Rank: 500 Posts; Posts: 148; Joined: Wed Dec 10, 2008 5:13 pm; Location: SF, CA; Thanked: 12 times

by lilu » Tue Mar 03, 2009 10:34 pm

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?
5/21
3/7
4/7
5/7
16/21

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Source: Beat The GMAT — Problem Solving | Tue Mar 03, 2009 10:34 pm

piyush_nitt: Master | Next Rank: 500 Posts; Posts: 424; Joined: Sun Dec 07, 2008 5:15 pm; Location: Sydney; Thanked: 12 times

Re: Probability-siblings

by piyush_nitt » Wed Mar 04, 2009 3:16 am

lilu wrote: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?
5/21
3/7
4/7
5/7
16/21

IMO E

There are 2 scenarios here :

1. Select 2 persons from group of 4 ,

Prob of selecting 2 person from group of 4 = 2C1 * 2C1 = 4

2. Select 1 from group of 4 and 1 from group of 3

Prob of selecting 1 person from group of 4 and 1 from group of 3 = 4C1*3C1 = 12

Total prob = 7C2 = 21

Therefore Ans is, Case 1 + Case 2 /21

16/21

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venmic: Master | Next Rank: 500 Posts; Posts: 222; Joined: Mon Oct 13, 2008 4:04 pmFollowed by: 2 members; Thanked: 3 times

by venmic » Wed Apr 27, 2011 10:29 pm

can anyone give a better explanation for thsi please...

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manpsingh87: Master | Next Rank: 500 Posts; Posts: 436; Joined: Tue Feb 08, 2011 3:07 amFollowed by: 6 members; Thanked: 72 times

by manpsingh87 » Wed Apr 27, 2011 11:18 pm

lilu wrote: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?
5/21
3/7
4/7
5/7
16/21

probability of not selecting siblings= 1- (probability of selecting siblings);
we have 7 people 4 of which have exactly 1 sibling and 3 have exactly 2 siblings..!!!
let 7 people be p1,p2,p3,p4,p5,p6,p7; now out of these 7 peoples 4 people having exactly 1 sibling can be (p1&p2), (p3&p4); and people having exaclty two siblings can be (p5,p6,p7).

lets call them group 1(p1&2), group2(p3&p4) and group 3(p5,p6,p7)
out of 7 people any two people can be selected in 7C2 ways=21;

probability that two persons selected belongs to group1 =2C2=1;
probability that two persons selected belongs to group 2=2C2=1;
probability that two persons selected belongs to group 3=3C2=3;

hence the probability that selected persons will be sibling= 1+1+3/21=5/21;

required probability= 1-5/21=16/21; hence E

O Excellence... my search for you is on... you can be far.. but not beyond my reach!

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Prashant Ranjan: Senior | Next Rank: 100 Posts; Posts: 40; Joined: Thu Jan 21, 2010 1:36 amFollowed by: 1 member; Thanked: 6 times

by Prashant Ranjan » Fri Jul 01, 2011 3:05 am

Consider the following sibling pairs S1, S2, and S3
S1 (A,B) S2(C,D) S3(E,F,G)

Now the probability of selecting A is 1C1. If A is selected then B cannot be selected and every other person can be selected. Thus the pairs can be AB, AC, AD, AE, AF, AG where the order of the selection doesn't matter. So the probability of this selection is 1C1 * 5C1 / 7C2.
In the same way, B can also be paired with 5 other persons but not A.
BC, BD, BE, BF, BG. So the selection probability is 1C1*5C1/7C2.
Now consider the selection of C. CE, CF, CG i.e. 1C1*3C1/7C2 (since CA and CB have already been counted).
Same with D. DE, DF, DG i.e. 1C1*3C1/7C2 (Since DA, DB have already been counted).
However when we select E we cannot select the persons F and G, since they are kins. Also we cannot select any other person from the 4 people since that pair has already been counted above.
Same goes for F and G. Thus the total probability of selection is
5C1/7C2 + 5C1/7C2 + 3C1/7C2 + 3C1/7C2 = 16C1/7C2 = 16/21

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