Question

11. Five statements $S_1,S_2,S_3,S_4,S_5$ are allotted specific seats $P_1,P_2,P_3,P_4,P_5$ to sit in the class. On the examination day they are allotted seats randomly. Let event $E_i$ be that $S_i$ and $S_{i+1}$ are not adjacent then find the value of $P(E_1\cap E_2\cap E_3\cap E_4)$.

Answer 1

Given, there are $5$ students $S_1,S_2,S_3,S_4,S_5$ and five seats $P_1,P_2,P_3,P_4,P_5$. Now, on the exam day students are alloted seats randomly such that $Si$ and $S_{iH}$ are not adjacent. Now,

$E_1$ means $S_1\&S_2$ do not seat together

$E_2$ means $S_2$ & $S_3$ do not sit together

$E_3$ means $S_3$ & $S_4$ do not sit together

$E_4$ means $S_4$ & $S_5$ do not sit together

So, we can see that $S_1$ & $S_5$ can not sit will only $S_2$ & $S_4$ respectively

Whereas $S_2$ & $S_3$ cannot sit with $(S_1,S_3)$ and $(S_2,S_4)$ respectively and $S_3$ we have to look separately

So, total ways in which we can randomly arrange $5$ students is $120\equiv (5!)$

Say, $S_1$ is on the first seat, then following are possible arrangement

So, similarly we will have two cases for $S_5$ as well

Now, say $S_2$ is on the first seat, the following are possible arrangement

Similarly for $S_4$ we will have $3$ possibilities

Now, say $S_3$ is on the first seat, the following are possible arrangement.

$\therefore$ $P(E_1\cap E_2\cap E_3\cap E_4)=\frac{2+2+3+3+4}{120}=\frac{14}{120}=\frac{7}{60}$