Question

There are \[n\] persons sitting in a row. Two of them are selected at random. The probability that the two selected persons are not together, is
A) \[\dfrac{2}{n}\]
B) \[1 - \dfrac{2}{n}\]
C) \[\dfrac{{n\left( {n - 1} \right)}}{{\left( {n + 1} \right)\left( {n + 2} \right)}}\]
D) None of these

Answer 1

Hint:
In this question, we need to find the probability of an event happening. Since 2 persons are ‘selected’ rather than ‘arranged’, hence, we would apply the formula of combinations. We will first find the number of ways of selecting 2 people at random. Then we will find the probability of the two selected persons sitting together. We will then simplify it further and subtract the obtained probability from 1 to get the probability that the two selected persons are not together.

Complete step by step solution:
Total number of people sitting in a row \[\]
Number of ways of selecting 2 of them at random \[ = {}^n{C_2}\]
Number of ways of selecting 2 persons sitting together= \[n - 1\]
Now, the probability of the two selected persons sitting together \[ = \] Two persons sitting together \[ \div \] Total number of ways of selecting 2 persons.
 \[ \Rightarrow \] Probability of the two selected persons sitting together \[ = \dfrac{{n - 1}}{{{}^n{C_2}}}\]
Now applying the formula of \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\], we get
 \[ \Rightarrow \] Probability of the two selected persons sitting together \[ = \dfrac{{n - 1}}{{\dfrac{{n!}}{{2!\left( {n - 2} \right)!}}}}\]
 \[ \Rightarrow \] Probability of the two selected persons sitting together \[ = \dfrac{{n - 1}}{{\dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)!}}{{2!\left( {n - 2} \right)!}}}}\]
Now solving further, we get,
 \[ \Rightarrow \] Probability of the two selected persons sitting together \[ = \dfrac{2}{n}\]
Now, as we know the total probability of an event is 1.
Therefore, the probabilities that the two selected persons are not together \[ = 1 - \dfrac{2}{n}\]

Hence, option B is the correct option.

Note:
An alternate way to solve this question is to directly find the probability of two selected persons not sitting together.
Number of ways in which 2 selected persons are not together \[ = {}^n{C_2} - \left( {n - 1} \right)\]
Therefore, the probability that the 2 selected persons are not together \[ = \dfrac{{{}^n{C_2} - \left( {n - 1} \right)}}{{{}^n{C_2}}}\]
 \[ \Rightarrow \] Probability of the two selected persons sitting together \[ = 1 - \dfrac{{\left( {n - 1} \right)}}{{{}^n{C_2}}}\]
Now, simplifying the expression using the formula of combination, we get
 \[ \Rightarrow \] Probability of the two selected persons sitting together \[ = 1 - \dfrac{2}{n}\]
Also, since, we had to do selection we used the formula of combination. If in the same question we had to do arrangement, we would have used the formula of permutation.