Mathematics · Probability

JEE Advanced 2018 — Paper 1 — Question 48

There are five students S1,S2,S3,S4S_{1}, S_{2}, S_{3}, S_{4} and S5S_{5} in a music class and for them there are five seats R1,R2,R3,R4R_{1}, R_{2}, R_{3}, R_{4} and R5R_{5} arranged in a row, where initially the seat RiR_{i} is allotted to the student Si,i=1,2,3,4,5S_{i}, i=1,2,3,4,5. But, on the examination day, the five students are randomly allotted the five seats.

For i=1, 2, 3, 4, let Ti_i denote the event that the students Si_i and Si+1_{i+1} do NOT sit adjacent to each other on the day of the examination. Then, the probability of the event T1∩_1 \cap T2∩_2 \cap T3∩_3 \cap T4_4 is

  1. Option A:

    115\frac{1}{15}

  2. Option B:

    110\frac{1}{10}

  3. Option C:

    760\frac{7}{60}

    Correct
  4. Option D:

    15\frac{1}{5}

Answer: C

Step-by-step solution

S1 S3 S5 S2 S4S_1 \ S_3 \ S_5 \ S_2 \ S_4 S2 S4 S1 S3 S5S_2 \ S_4 \ S_1 \ S_3 \ S_5 S3 S5 S1 S4 S2S_3 \ S_5 \ S_1 \ S_4 \ S_2 S3 S5 S2 S4 S1S_3 \ S_5 \ S_2 \ S_4 \ S_1 S4 S1 S3 S5 S2S_4 \ S_1 \ S_3 \ S_5 \ S_2 S4 S2 S5 S1 S3S_4 \ S_2 \ S_5 \ S_1 \ S_3 S5 S2 S4 S1 S3S_5 \ S_2 \ S_4 \ S_1 \ S_3

Same number of ways in reverse order

P(E)P(E) = N(E)5!=7×25!=760\frac{N(E)}{5!} = \frac{7 \times 2}{5!} = \frac{7}{60}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2018
Paper
Paper 1
Subject
Mathematics
Chapter
Probability
Topic
Addition Theorem, Venn Diagrams and Types of Events
There are five students S 1 , S 2 , S 3 , S 4 and S 5 in a music… | JEE Advanced 2018 PYQ with Solution · DhiX AI