Question

Suppose $n \geq 3$ persons are sitting in a row. Two of them are selected at random. The probability that they are not together is

• A. $1 - \frac { 2 } { n }$
• B. $\frac { 2 } { n - 1 }$
• C. $1 - \frac { 1 } { n }$
• D. None of these

Answer 1

Answer: (A)

$1 - \frac { 2 } { n }$

Answer 2

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

$X \bullet X \bullet X \bullet X \bullet X \bullet X$

$P = \frac { { } ^ { n - 1 } C _ { 2 } } { { } ^ { n } C _ { 2 } } = \frac { ( n - 1 ) ! 2 ! ( n - 2 ) ! } { ( n - 3 ) ! 2 ! n ! } = \frac { n - 2 } { n } = 1 - \frac { 2 } { n }$ .