JEE Advance - Mathematics (2018 - Paper 1 Offline - No. 18)
There are five students S1, S2, S3, S4 and S5 in a music class and for them there are five seats R1, R2, R3, R4 and R5 arranged in a row, where initially the seat Ri is allotted to the student Si, i = 1, 2, 3, 4, 5. But, on the examination day, the five students are randomly allotted the five seats.
(There are two questions based on Paragraph "A", the question given below is one of them)
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 $${T_1} \cap {T_2} \cap {T_3} \cap {T_4}$$ is
(There are two questions based on Paragraph "A", the question given below is one of them)
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 $${T_1} \cap {T_2} \cap {T_3} \cap {T_4}$$ is
$${1 \over {15}}$$
$${1 \over {10}}$$
$${7 \over {60}}$$
$${1 \over {5}}$$
Explanation
Here, $$n({T_1} \cap {T_2} \cap {T_3} \cap {T_4})$$ = Total $$ - n(\overline {{T_1}} \cup \overline {{T_2}} \cup \overline {{T_3}} \cup \overline {{T_4}} )$$
$$ \Rightarrow $$ $$n({T_1} \cap {T_2} \cap {T_3} \cap {T_4})$$
$$ = 5! - [{}^4{C_1}4!2!\, - ({}^3{C_1}3!2! + {}^3{C_1}3!2!2!) + ({}^2{C_1}2!2!\, + \,{}^4{C_1}2.2!) - 2]$$
$$ \Rightarrow $$ $$n({T_1} \cap {T_2} \cap {T_3} \cap {T_4})$$
$$ = 120 - [192 - (36 + 72) + (8 + 16) - 2]$$
$$ = 120 - [192 - 108 + 24 - 2] = 14$$
$$ \therefore $$ Required probability = $${{14} \over {120}} = {7 \over {60}}$$
$$ \Rightarrow $$ $$n({T_1} \cap {T_2} \cap {T_3} \cap {T_4})$$
$$ = 5! - [{}^4{C_1}4!2!\, - ({}^3{C_1}3!2! + {}^3{C_1}3!2!2!) + ({}^2{C_1}2!2!\, + \,{}^4{C_1}2.2!) - 2]$$
$$ \Rightarrow $$ $$n({T_1} \cap {T_2} \cap {T_3} \cap {T_4})$$
$$ = 120 - [192 - (36 + 72) + (8 + 16) - 2]$$
$$ = 120 - [192 - 108 + 24 - 2] = 14$$
$$ \therefore $$ Required probability = $${{14} \over {120}} = {7 \over {60}}$$
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