InterviewSolution
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InterviewSolution
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There are five students `S_(1),S_(2),S_(3),S_(4)` and `R_(5)` arranged in a row, where initially the seat `R_(1)`is allotted to the students are randomly allotted the five seats `R_(1),R_(2),R_(3),R_(4)` and `R_(6)` 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. (There are two questions based on Paragraph, the question given below is one of them) For i=1,2,3,4. let `T_(i)` denote the event that the students `S_(1)` 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)` isA. `(1)/(15)`B. `(1)/(10)`C. `(7)/(60)`D. `(1)/(6)` |
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Answer» Correct Answer - C Here , ` n(T_(1) cap T_(2) cap T_(3) cap T_(4))` Total `=-n(bar(T_(1)) cup bar(T_(2)) cup bar(T_(3)) cup bar(T_(4))) ` `implies 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]` ` implies 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` `:.` Required probability `=(14)/(120)=(7)/(60)` |
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