SATHEE: Probability 2 Question 17

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) $\frac{1}{10}$

(d) $\frac{1}{5}$

Show Answer

Solution:

Here, $n (T_{1} \cap T_{2} \cap T_{3} \cap T_{4})$

Total $= - n (\overset{―}{T_{1}} \cup \overset{―}{T_{2}} \cup \overset{―}{T_{3}} \cup \overset{―}{T_{4}})$

$\Rightarrow n (T_{1} \cap T_{2} \cap T_{3} \cap T_{4})$

$\begin{aligned} = 5! - [^{4} C_{1} 4! 2! - (^{3} C_{1} 3! 2! +^{3} C_{1} 3! 2! 2!) \\ + (^{2} C_{1} 2! 2! +^{4} C_{1} 2 \cdot 2!) - 2] \\ \Rightarrow n (T_{1} \cap T_{2} \cap T_{3} \cap T_{4}) \\ = 120 - [192 - (36 + 72) + (8 + 16) - 2] \\ = 120 - [192 - 108 + 24 - 2] = 14 \\ ∴ Required probability = \frac{14}{120} = \frac{7}{60} \end{aligned}$

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Probability 2 Question 17

17. For i=1,2,3,4, let Ti denote the event that the students Si and Si+1 do NOT sit adjacent to each other on the day of the examination. Then, the probability of the event T1∩T2∩T3∩T4 is

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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