Problem5.35 Solutionfromeq.(5.2) published presentations and documents on DocSlides.
b+c+1+b c+a+1+c a+b+1+(1 a)(1 b)(1 c)1:Problem4(U...
Problem5.35 Solution: FromEq.(5.2),1=B1 m1=1 m1(
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Pnk=1X2k1=2! ;indistribution,where isthe...
1pforjpj1;wewouldget1Xk=0kpk1=1 (1p)2;...
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