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

Hint: 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{align}
  & {{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{align} $
The above arrangement can be reversed and arranged again.
So, we get,
 $ \begin{align}
  & 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{align} $
Let us now calculate $ N $ , that is, the total number of sitting arrangements in the examination.
As there were five students, we get,
 $ \begin{align}
  & N=5! \\
 & \Rightarrow N=5\times 4\times 3\times 2\times 1 \\
 & \Rightarrow N=120 \\
\end{align} $
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{align}
  & 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{align} $
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{align}
  & 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{align} $
So, while arranging the objects we should remember to include the arrangements that are in the reverse of the order we have taken too.