Probability Question 429
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
Correct Answer: AShow Answer
Answer:
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} $ .
BETA
