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PROB ability....

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by Nigogo » Mon Oct 12, 2009 7:08 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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by nithi_mystics » Mon Oct 12, 2009 7:27 pm
Out of 7, we have 2 pairs of siblings and 3 siblings (2+2+3).

Probability that the siblings from the 1st pair are selected = 2C2/7C2

Probability that the siblings from the 2nd pair are selected = 2C2/7C2

Probability that 2 from the 3 siblings are selected = 3C2/7C2

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

Probability that the ones selected are not siblings
= 1- 5/21
= 16/21

Whats the OA?
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by nithi_mystics » Wed Oct 14, 2009 8:28 am
OA plz
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by Nigogo » Wed Oct 14, 2009 9:10 pm
different way but the same answer
We are told that 4 people have exactly 1 sibling. This would account for 2 sibling relationships (e.g. AB and CD). We are also told that 3 people have exactly 2 siblings. This would account for another 3 sibling relationships (e.g. EF, EG, and FG). Thus, there are 5 total sibling relationships in the group.

Additionally, there are (7 x 6)/2 = 21 different ways to chose two people from the room.

Therefore, the probability that any 2 individuals in the group are siblings is 5/21. The probability that any 2 individuals in the group are NOT siblings = 1 - 5/21 = 16/21.

The correct answer is E. :!: :arrow: :arrow: :roll:
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