Start free

Mathematics · Probability

JEE Advanced 2018 — Paper 1 — Question 47

The probability that, on the examination day, the student S1_1 gets the previously allotted seat R1_1, and NONE of the remaining students gets the seat previously allotted to him/her is

  1. Option A:

    340\frac{3}{40}

    Correct
  2. Option B:

    18\frac{1}{8}

  3. Option C:

    740\frac{7}{40}

  4. Option D:

    15\frac{1}{5}

Answer: A

Step-by-step solution

Total number of ways to seat 5 students in 5 seats = 5!=1205! = 120. We want exactly one fixed point: student S1S_1 gets seat R1R_1. The remaining 4 students must form a derangement (no one gets their own seat). Number of derangements of 4 items, !4=D4=9!4 = D_4 = 9. Favorable permutations = 1×9=91 \times 9 = 9. Required probability = 9120=340\dfrac{9}{120} = \dfrac{3}{40}.

Answer key and solution verified before publishing.

Practise Probability

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
JEE Advanced 2018
Paper
Paper 1
Subject
Mathematics
Chapter
Probability
Topic
Addition Theorem, Venn Diagrams and Types of Events

More Probability questions from this paper

The probability that, on the examination day, the student S 1 gets… | JEE Advanced 2018 PYQ with Solution · DhiX AI