Question

For i = 1,2,3,4 let ${{T}_{i}}$ denote the event that 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 ${{T}_{1}}\bigcap {{T}_{2}}\bigcap {{T}_{3}}\bigcap {{T}_{4}}$ is?

Answer 1

We start solving the given problem by considering the event ${{T}_{i}}$ that the students ${{S}_{i}}$ and ${{S}_{i+1}}$ do not sit adjacent to each other on the day of the examination for i = 1,2,3,4 . Then we consider the all possible outcomes for the event ${{T}_{i}}$ , and then we find the probability of the event ${{T}_{i}}$ using the formula $P\left( E \right)=\dfrac{n\left( E \right)}{n\left( S \right)}$ . Hence we get the final answer.

Complete step-by-step answer:
Let us consider the event ${{T}_{i}}$ that the students ${{S}_{i}}$ and ${{S}_{i+1}}$ do not sit adjacent to each other on the day of the examination for i = 1,2,3,4 .
We consider the event ${{T}_{1}}$ that the students ${{S}_{1}}$ and ${{S}_{2}}$ do not sit adjacent to each other on the day of the examination.
We consider the event ${{T}_{2}}$ that the students ${{S}_{2}}$ and ${{S}_{3}}$ do not sit adjacent to each other on the day of the examination.
We consider the event ${{T}_{3}}$ that the students ${{S}_{3}}$ and ${{S}_{4}}$ do not sit adjacent to each other on the day of the examination.
We consider the event ${{T}_{4}}$ that the students ${{S}_{4}}$ and ${{S}_{5}}$ do not sit adjacent to each other on the day of the examination.
Now, let us consider the possible outcomes for the event ${{T}_{1}}\bigcap {{T}_{2}}\bigcap {{T}_{3}}\bigcap {{T}_{4}}$ , we get,
$\begin{aligned} & {{S}_{1}}{{S}_{3}}{{S}_{5}}{{S}_{2}}{{S}_{4}} \\\ & {{S}_{2}}{{S}_{4}}{{S}_{1}}{{S}_{3}}{{S}_{5}} \\\ & {{S}_{3}}{{S}_{5}}{{S}_{1}}{{S}_{4}}{{S}_{2}} \\\ & {{S}_{3}}{{S}_{5}}{{S}_{2}}{{S}_{4}}{{S}_{1}} \\\ & {{S}_{4}}{{S}_{1}}{{S}_{3}}{{S}_{5}}{{S}_{2}} \\\ & {{S}_{4}}{{S}_{2}}{{S}_{5}}{{S}_{1}}{{S}_{3}} \\\ & {{S}_{5}}{{S}_{2}}{{S}_{4}}{{S}_{1}}{{S}_{3}} \\\ \end{aligned}$
The above arrangement can be reversed and arranged again.
So, we get,
$\begin{aligned} & n\left( {{T}_{1}}\bigcap {{T}_{2}}\bigcap {{T}_{3}}\bigcap {{T}_{4}} \right)=2\times 7 \\\ & \Rightarrow n\left( {{T}_{1}}\bigcap {{T}_{2}}\bigcap {{T}_{3}}\bigcap {{T}_{4}} \right)=14 \\\ \end{aligned}$
Let us now calculate $N$ , that is, the total number of sitting arrangements in the examination.
As there were five students, we get,
$\begin{aligned} & N=5! \\\ & \Rightarrow N=5\times 4\times 3\times 2\times 1 \\\ & \Rightarrow N=120 \\\ \end{aligned}$
Consider the formula, probability of occurring of an event $E$ is $P\left( E \right)=\dfrac{n\left( E \right)}{n\left( S \right)}$
By using the above formula, we get,
$\begin{aligned} & P\left( {{T}_{1}}\bigcap {{T}_{2}}\bigcap {{T}_{3}}\bigcap {{T}_{4}} \right)=\dfrac{n\left( {{T}_{1}}\bigcap {{T}_{2}}\bigcap {{T}_{3}}\bigcap {{T}_{4}} \right)}{N} \\\ & \Rightarrow P\left( {{T}_{1}}\bigcap {{T}_{2}}\bigcap {{T}_{3}}\bigcap {{T}_{4}} \right)=\dfrac{14}{120} \\\ & \Rightarrow P\left( {{T}_{1}}\bigcap {{T}_{2}}\bigcap {{T}_{3}}\bigcap {{T}_{4}} \right)=\dfrac{7}{60} \\\ \end{aligned}$
Therefore, the required probability is $P\left( {{T}_{1}}\bigcap {{T}_{2}}\bigcap {{T}_{3}}\bigcap {{T}_{4}} \right)=\dfrac{7}{60}$
Hence, the answer is $\dfrac{7}{60}$ .

Note: The possibility of making a mistake in this problem is one might not consider the reverse arrangement of the students in the examination hall, if we do this mistake, we get
$\begin{aligned} & P\left( {{T}_{1}}\bigcap {{T}_{2}}\bigcap {{T}_{3}}\bigcap {{T}_{4}} \right)=\dfrac{n\left( {{T}_{1}}\bigcap {{T}_{2}}\bigcap {{T}_{3}}\bigcap {{T}_{4}} \right)}{N} \\\ & \Rightarrow P\left( {{T}_{1}}\bigcap {{T}_{2}}\bigcap {{T}_{3}}\bigcap {{T}_{4}} \right)=\dfrac{7}{120} \\\ \end{aligned}$
So, while arranging the objects we should remember to include the arrangements that are in the reverse of the order we have taken too.