Subject: Mathematics — Probability
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
Answer: $\frac{7}{60}$
Solution: Total cases = $5!$.
Let us count the number of favourable ways.
$$\begin{matrix}
2 & 4 & 1 & \times & \times \\
2 & 5 & 3 & 1 & 4
\end{matrix}\quad\} 3\text{ ways} \times 2 = 6$$
$$\begin{matrix}
1 & 3 & 5 & 2 & 4 \\
1 & 4 & 2 & 5 & 3
\end{matrix}\quad\} 2\text{ ways} \times 2 = 4$$
$$\begin{matrix}
3 & 1 & 5 & 2 & 4 \\
3 & 1 & 4 & 2 & 5
\end{matrix}\quad\} 2\text{ ways} \times 2 = 4$$
$$\therefore\text{Total favourable ways} = 14.$$
Alternative Solution:
(By P.I.E.)
$$\left| T_{1} \cap T_{2} \cap T_{3} \cap T_{4} \right| = |T| - \left| T_{1}' \cup T_{2}' \cup T_{3}' \cup T_{4}' \right|$$
$$N = 5! - \left\{ \binom{4}{1} \cdot 4! \cdot 2! - \left( \binom{3}{1} \cdot 3! \cdot 2! + \binom{3}{1} \cdot 3! \cdot 2! \cdot 2! \right) + \left( \binom{2}{1} \cdot 2! \cdot 2! + \binom{4}{1} \cdot 2! \cdot 2! \right) - 2 \right\}$$
$$N = 5! - \{ 4 \cdot 48 - (3 \cdot 6 \cdot 2 + 3 \cdot 6 \cdot 2 \cdot 2) + (2 \cdot 2 \cdot 2 + 4 \cdot 2 \cdot 2) - 2\}$$
$$N = 120 - (192 - 108 + 24 - 2) = 120 - 106 = 14.$$
Hence, the required probability
$$\text{Required probability} = \frac{14}{120} = \frac{7}{60}.$$
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