Probability Question 126
Question: Suppose $ n\ge 3 $ persons are sitting in a row. Two of them are selected at random. The probability that they are not together is
[Pb. CET 2004]
Options:
A) $ 1-\frac{2}{n} $
B) $ \frac{2}{n-1} $
C) $ 1-\frac{1}{n} $
D) None of these
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Answer:
Correct Answer: A
Solution:
Let there be n persons and $ (n-2) $ persons not selected are arranged in places stated above by stars and the selected 2 persons can be arranged at places stated by dots (dots are $ n-1 $ in number) So the favourable ways are $ ^{n-1}C_2 $ and the total ways are $ ^{n}C_2 $ , so $ \times \bullet \times \bullet \times \bullet \times \bullet \times \bullet \times $
$ P=\frac{^{n-1}C_2}{^{n}C_2}=\frac{(n-1)!2!(n-2)!}{(n-3)!2!n!}=\frac{n-2}{n}=1-\frac{2}{n} $ .
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