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JEE Advanced 2018Paper 1
Exam
JEE Advanced
Year
2018
Paper
Paper 1
Subject
Mathematics
Chapter
Probability

JEE Advanced 2018 · Paper 1 · Mathematics · Probability

Paragraph: There are five students S _ 1 , S _ 2 , S _ 3 , S _ 4...

Paragraph: There are five students S1,S2,S3,S4{S}_{1}, {S}_{2}, {S}_{3}, {S}_{4} and S5{S}_{5} in a music class and for them there are five seats R1,R2,R3,R4{R}_{1}, {R}_{2}, {R}_{3}, {R}_{4} and R5{R}_{5} arranged in a row, where initially the seat Ri{R}_{i} is allotted to the student Si,i=1,2,3,4,5.{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=1,2,3,4i = 1, 2, 3, 4 Let Ti{T}_{i} denote the event that the students Si{S}_{i} and Si+1{S}_{i + 1} do NOT sit adjacent to each other on the day of the examination. Then, the probability of the event T1∩T2∩T3∩T4{T}_{1} \cap {T}_{2} \cap {T}_{3} \cap {T}_{4} is

Options

  1. A115\dfrac{1}{15}
  2. B110\dfrac{1}{10}
  3. C760\dfrac{7}{60}
  4. D15\dfrac{1}{5}

Correct answer

C. 760\dfrac{7}{60}

Step-by-step solution

n(T1∩T2∩T3∩T4)=Total−n(T1¯∪T2¯∪T3¯∪T4¯)n\left({T}_{1} \cap {T}_{2} \cap {T}_{3} \cap {T}_{4}\right) = Total - n\left(\overset{¯}{{T}_{1}} \cup \overset{¯}{{T}_{2}} \cup \overset{¯}{{T}_{3}} \cup \overset{¯}{{T}_{4}}\right)
=5!−(4C14!2!−(3C1.3!2!+3C13!2!2!)+(2C12!2!+4C1.2.2!)−2)= 5 ! - \left({}^{4}{C}_{1}4 ! 2 ! - \left({}^{3}{C}_{1} . 3 ! 2 ! + {}^{3}{C}_{1}3 ! 2 ! 2 !\right) + \left({}^{2}{C}_{1}2 ! 2 ! + {}^{4}{C}_{1} . 2.2 !\right) - 2\right)
=14= 14
Probability =145!=760= \dfrac{14}{5 !} = \dfrac{7}{60}
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