Probability Question

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chrisjim5: Newbie | Next Rank: 10 Posts; Posts: 8; Joined: Wed Oct 06, 2010 11:35 am; Thanked: 1 times

by chrisjim5 » Sun Oct 24, 2010 8:56 am

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?

Can you please provide an answer to this?

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Source: Beat The GMAT — Problem Solving | Sun Oct 24, 2010 8:56 am

shovan85: Community Manager; Posts: 991; Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members; Location: Bangalore, India; Thanked: 146 times

by shovan85 » Sun Oct 24, 2010 9:23 am

chrisjim5 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?

Can you please provide an answer to this?

Let's the 7 people be ABCDEFG.

4 people have exactly 1 sibling:

Let A and B are siblings and C and D are siblings.

So,
A has 1 sibling B
B has 1 sibling A
C has 1 sibling D
D has 1 sibling C

3 people have exactly 2 siblings:

Let E, F and G are all siblings of each other.

So,
E has 2 siblings (F and G).
F has 2 siblings (E and G).
G has 2 siblings (E and F).

Total number of sibling pairs = 5 [AB, CD, EF, EG, FG]

Total number of pairs that can be formed from 7 people C(7,2) = 7!/5!*2! = 21

P(sibling pair) = 5/21

P(not sibling pair) = 1 - 5/21 = 16/21.

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waltz2salsa: Junior | Next Rank: 30 Posts; Posts: 28; Joined: Tue Sep 28, 2010 9:19 amFollowed by: 1 member; Thanked: 3 times

by waltz2salsa » Thu Oct 28, 2010 11:25 pm

Can someone please tell me where am i going wrong!!!

7 people,

4 have 1 siblings i.e. (a1, a2) and (b1, b2)
3 have 2 siblings i.e. (c1,c2,c3)

total ways of selecting two people out of 7 = 7c2

favorable ways:

(ways of selecting one from Ci *(ways of selecting one from left 4people) + (ways of selecting one from Ai *(ways of selecting one from left 5 people) + (ways of selecting one from Bi *(ways of selecting one from left 5 people)

= 3c1*4c1 + 2c1*5c1 + 2c1*5c1

P(no siblings) = fav ways/total ways = (3c1*4c1 + 2c1*5c1 + 2c1*5c1) / 7c2

= 2( 16/21)

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Fri Oct 29, 2010 12:50 am

waltz2salsa wrote:Can someone please tell me where am i going wrong!!!

7 people,

4 have 1 siblings i.e. (a1, a2) and (b1, b2)
3 have 2 siblings i.e. (c1,c2,c3)

total ways of selecting two people out of 7 = 7c2

favorable ways:

(ways of selecting one from Ci *(ways of selecting one from left 4people) + (ways of selecting one from Ai *(ways of selecting one from left 5 people) + (ways of selecting one from Bi *(ways of selecting one from left 5 people)

= 3c1*4c1 + 2c1*5c1 + 2c1*5c1

P(no siblings) = fav ways/total ways = (3c1*4c1 + 2c1*5c1 + 2c1*5c1) / 7c2

= 2( 16/21)

You're overcounting the number of non-sibling pairs:

Number of ways to combine 1 choice from (c1,c2,c3) and 1 choice from (a1,a2,b1,b2) = 3*4 = 12.

We've now counted all the non-sibling pairs that include (c1, c2, c3). We need to count only the non-sibling pairs that don't include (c1,c2,c3).
Number of ways to combine 1 choice from (a1,a2) and 1 choice from (b1,b2) = 2*2 = 4.

Total non-sibling pairs = 12+4 = 16.

Total possible pairs from 7 people = 7C2 = 21.

P(non-sibling pair) = 16/21.

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