Probability 2 Question 17
17. 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) $\frac{1}{5}$
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Solution:
- Here, $n\left(T _1 \cap T _2 \cap T _3 \cap T _4\right)$
Total $=-n\left(\overline{T _1} \cup \overline{T _2} \cup \overline{T _3} \cup \overline{T _4}\right)$
$\Rightarrow n\left(T _1 \cap T _2 \cap T _3 \cap T _4\right)$
$$ \begin{aligned} & =5 !-\left[{ }^{4} C _1 4 ! 2 !-\left({ }^{3} C _1 3 ! 2 !+{ }^{3} C _1 3 ! 2 ! 2 !\right)\right. \\ & \left.+\left({ }^{2} C _1 2 ! 2 !+{ }^{4} C _1 2 \cdot 2 !\right)-2\right] \\ & \Rightarrow \quad n\left(T _1 \cap T _2 \cap T _3 \cap T _4\right) \\ & =120-[192-(36+72)+(8+16)-2] \\ & =120-[192-108+24-2]=14 \\ & \therefore \quad \text { Required probability }=\frac{14}{120}=\frac{7}{60} \end{aligned} $$
