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Question: There are \[n\] persons sitting in a row. Two of them are selected at random. The probability that t...

There are nn persons sitting in a row. Two of them are selected at random. The probability that the two selected persons are not together, is
A) 2n\dfrac{2}{n}
B) 1−2n1 - \dfrac{2}{n}
C) n(n−1)(n+1)(n+2)\dfrac{{n\left( {n - 1} \right)}}{{\left( {n + 1} \right)\left( {n + 2} \right)}}
D) None of these

Explanation

Solution

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 =nC2 = {}^n{C_2}
Number of ways of selecting 2 persons sitting together= n−1n - 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 =n−1nC2 = \dfrac{{n - 1}}{{{}^n{C_2}}}
Now applying the formula of nCr=n!r!(n−r)!{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}, we get
⇒\Rightarrow Probability of the two selected persons sitting together =n−1n!2!(n−2)! = \dfrac{{n - 1}}{{\dfrac{{n!}}{{2!\left( {n - 2} \right)!}}}}
⇒\Rightarrow Probability of the two selected persons sitting together =n−1n(n−1)(n−2)!2!(n−2)! = \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 =2n = \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 =1−2n = 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 =nC2−(n−1) = {}^n{C_2} - \left( {n - 1} \right)
Therefore, the probability that the 2 selected persons are not together =nC2−(n−1)nC2 = \dfrac{{{}^n{C_2} - \left( {n - 1} \right)}}{{{}^n{C_2}}}
⇒\Rightarrow Probability of the two selected persons sitting together =1−(n−1)nC2 = 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 =1−2n = 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.