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In a room filled with 7 people, 4 people have exactly 1 sibl [#permalink]

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05 Jul 2010, 13:45

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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?

Determine how many pairs of siblings there are. There are 5. 1 from G1, 1 from G2, 3 from G3. So the probability that two people are siblings to be picked from 7 people is 5/21.

Here's the explanation. The fact that 4 people have exactly 1 sibling means that these four people can be grouped into two. P1 is sibling with P2, and P3 with P4. These four people have only one sibling.

The fact that 3 people have two siblings mean that all three are siblings, with one person from this group related to the two other. So P5 is related to P6 and P7. These three people have two siblings each.

All combinations for choosing in this group can be noted as 7!/5!2! = 21

Now, we have three groups of siblings. We must choose two people that are not siblings, to complete the probability. We must do this in 3 cases. 3!/2!1! = 3. Two draws, one per group, from three groups. G1 to G2, G1 to G3, G2 to G3.

So we'd have:

G1 to G2 - 2*2 = 4 G1 to G3 - 2*3 = 6 G2 to G3 - 2*3 = 6

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

As there are 4 people with exactly 1 sibling each: we have two pairs of siblings (1-2; 3-4). As there are 3 people with exactly 2 siblings each: we have one triple of siblings (4-5-6).

Solution #1: # of selections of 2 out of 7 - \(C^2_7=21\); # of selections of 2 people which are not siblings - \(C^1_2*C^1_2\) (one from first pair of siblings*one from second pair of siblings)+\(C^1_2*C^1_3\) (one from first pair of siblings*one from triple)+ \(C1^_2*C^1_3\)(one from second pair of siblings*one from triple) \(=4+6+6=16\).

\(P=\frac{16}{21}\)

Solution #2: # of selections of 2 out of 7 - \(C^2_7=21\); # of selections of 2 siblings - \(C^2_3+C^2_2+C^2_2=3+1+1=5\);

(4/7)*(5/6) + (3/7)*(4/6) = 16/21 Case 1: First person selected has one sibling -> prob is 4/7 , then second person selected must not be his/her sibling prob 5/6. Case 2: First person selected has two sibling -> prob is 3/7 , then second person selected must not be his/her sibling prob 4/6.
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