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Solution of probability

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  • "Ans. 1Possible arrangements- 20!1 1 1 1 1 P ? 1 ? ? ? ? .... ? ? ? 0.3679 1! 2! 3! 19! 20! E So, take as the event that no student is on the same seat , in relation to that, c E is atleast one student on same seat.The relation between the both event..

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  • "Ans. 1Possible arrangements- 20!1 1 1 1 1 P ? 1 ? ? ? ? .... ? ? ? 0.3679 1! 2! 3! 19! 20! E So, take as the event that no student is on the same seat , in relation to that, c E is atleast one student on same seat.The relation between the both events,c P(E)?? 1 P(E )th Same as, take A be the same event that thei student is on the same seat.i So, apply the summation rule for the all twenty students,20 c P(E ) ? P( A ) ? i i ?1 Since,t ?20 n P(A ) ? 1/ n; P(A ? A ) ? 1/ (n(n ?1) ;P( A ) ? 1/ ( )i!) i i j ?ii i ?1Hence, 20 20 1 c n i?? 11 i P(E ) ? ( ) ? ? ( ?1) ? 1/i!( ?1) ? 0.632 ?? i n ( )i! ii ?? 11 i PE ( ) ? 0.368 Hence Proved. "

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