| [ QuizWit ] in KIDS 글 쓴 이(By): random (김덕현 ) 날 짜 (Date): 1995년10월17일(화) 05시08분36초 KST 제 목(Title): re)사랑의 교차선 ... Pr(AUB) = Pr(A)+Pr(B)-Pr(AIB) U:union I:Intersection More generally, Pr(UAi)=Pr(A1UA2UA3U...UAN) =sum{Pr(Ai)}-sum{Pr(AiIAj),i<j}+sum{Pr(AiIAjIAk),i<j<k} +(-1)^(N-1)*Pr{A1IA2I...AN} Let the event Ai be the case when there is a match between a pair. Since there are n^2 possible single matches and Pr(Ai)=1/n^2, sum{Pr(Ai)}=1/n^2*n^2=1 There are {nC2}*{nP2} possible double-matches with Pr(AiIAj)=1/n^4, sum{Pr(AiIAj,i<j}=1/n^4*{nC2}{nP2} .... Now the maximum possible matches are n therefore Pr{A1IA2I...IAn}=1/n^{2n}*n! Then Pr(UAi)=1-1/n^4*{nC2}{nP2}+....+(-1)^(n-1)*1/n^{2n}*n!. We want probability of no matches. Pr(no matches)=1-Pr(some matches)=1-Pr(UAi) =sum[(-1)^i*n^(-2i)*{nCi}{nPi}],i=2..n ***{nPi} -> n permutation i {nCi} -> n combination i. I think this answer is the same as the Professor's one above. Anyone knows how to find the limit of this? How to find it to be 1/e... q |