JEE Advanced · Mathematics · 32. Probability

Paragraph: There are five students $S_{1}, S_{2}, S_{3}, S_{4}$ and $S_{5}$ in a music class and for them there are five seats $R_{1}, R_{2}, R_{3}, R_{4}$ and $R_{5}$ arranged in a row, where initially the seat $R_{i}$ is allotted to the student $S_{i}, i = 1,2, 3,4, 5 .$ But, on the examination day, the five students are randomly allotted the five seats
Question : 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 $\frac{1}{5}$

Verified Solution

Answer & Solution

Correct Answer

(C) $\frac{7}{60}$

Step-by-step Solution

Detailed explanation

n (T_{1} \cap T_{2} \cap T_{3} \cap T_{4}) = T o t a l - n (\bar{T_{1}} \cup \bar{T_{2}} \cup \bar{T_{3}} \cup \bar{T_{4}})

= 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} . 2.2!) - 2)

= 14

Probability

= \frac{14}{5!} = \frac{7}{60}

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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