Question

For $i=1,2,3,4$ let $T_i$ denote the event that the students $S_i$ and $S_{i+1}$ do NOT sit adjacent to each other on the day of the examination. Then the probability of the event $T_1\cap T_2\cap T_3\cap T_4$ is?

• A. $\frac{1}{15}$
• B. $\frac{1}{10}$
• C. $\frac{7}{60}$
• D. $\frac{1}{5}$

Answer 1

The correct option is C $\frac{7}{60}$

$n(T_1\cap T_2\cap T_3\cap T_4)=$Total $-n(\overline{T_1}\cup \overline{T_2}\cup \overline{T_3}\cup \overline{T_4})$
$=5!-({}^4{C}_{1}4!2!-({}^3{C}_1.3!2!+{}^3{C}_{1}3!2!2!)+({}^2{C}_{1}2!2!+{}^4{C}_1.2.2!)-2)$
$=14$
Probability $=\frac{14}{5!}=\frac{7}{60}$.