Question:
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:
C. 2/7
Explanation:
Given that 7 persons are to be seated in a row.
If two persons sit next to each other, then consider these two persons as 1 group.
Now we have to arrange 6 persons.
$\therefore$ Number of arrangement $=2 ! \times 6 !$
Total number of arrangement of 7 persons $=7 !$
Probability $=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}$
$\therefore$ Required probability $=\frac{2 ! \times 6 !}{7 !}$
$=\frac{2 \times 1 \times 6 !}{7 \times 6 !}$
$=\frac{2}{7}$
Hence, the correct option is (C).
= 2/7
Hence, the correct option is (C)