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15 May 2007, 02:35
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In a room filled with 7 people, 4 ppl have exactly 1 friend in room and 3 ppl have exactly 2 friends in room (Assuming that friendship is a mutual relationship). If 2 individuals are selected from the room at random, what is the probability that those 2 individuals are not friends.
A. 5/21
B. 3/7
C. 4/7
D. 5/7
E. 16/21
A. 5/21
B. 3/7
C. 4/7
D. 5/7
E. 16/21
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15 May 2007, 09:24
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I don't think 1-3c2/7c2 = 5/7. I'm getting 6/7...
Anyone else? Where am I going wrong?
(3!/2!1!)/(7!/2!5!) = (6/2)/(42/2) = 3/21 = 1/7
Therefore, 1-1/7 = 6/7.
I'm stuck...
Anyone else? Where am I going wrong?
(3!/2!1!)/(7!/2!5!) = (6/2)/(42/2) = 3/21 = 1/7
Therefore, 1-1/7 = 6/7.
I'm stuck...
