Choose the correct answer out of four given options in each of the exercise seven persons are to be seated in a row. The probability that two particular persons sit next to each other is A. $\frac{1}{3}$ B. $\frac{1}{6}$ C. $\frac{2}{7}$ D. $\frac{1}{2}$

Home » Choose the correct answer out of four given options in each of the exercise seven persons are to be seated in a row. The probability that two particular persons sit next to each other is A. B. C. D.

Choose the correct answer out of four given options in each of the exercise seven persons are to be seated in a row. The probability that two particular persons sit next to each other is
A. $\frac{1}{3}$
B. $\frac{1}{6}$
C. $\frac{2}{7}$
D. $\frac{1}{2}$

CBSE | Class 11 | Maths | NCERT Exemplar | Probability

Choose the correct answer out of four given options in each of the exercise seven persons are to be seated in a row. The probability that two particular persons sit next to each other is
A. $\frac{1}{3}$
B. $\frac{1}{6}$
C. $\frac{2}{7}$
D. $\frac{1}{2}$

Solution:

Option (C) $2 / 7$ is correct.
Explanation:
It is given that 7 persons are to be seated in a row.
If two persons sit next to each other, then they’ll be considered as 1 group.
We now have to arrange 6 persons.
$\therefore$ No. of arrangement $=2 ! \times 6 !$
Total no. of arrangement of 7 persons $=7 !$
$\begin{array}{l} \text { Probability }=\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }} \\ \therefore \text { Required probability }=\frac{2 ! \times 6 !}{7 !} \\ =\frac{2 \times 1 \times 6 !}{7 \times 6 !} \\ =\frac{2}{7} \end{array}$
As a result, the correct option is (C).