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11.[29 35]a unique common fixed point theorem under psi varphi contractive co...

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Let X be a geometric random variable with parameter p, and n and m be nonnegative integers. a. for what values of n is P(X=n) Maximum? b. what is the probability that X is even? Solution Let X be a geometric random variable with parameter p, and n and m be nonnegative integers. a. for what values of n is P(X=n) Maximum? P(x=n) = (1-p)^n-1 p we know that 0<=p<=1 now => 0<= 1- p <=1 (1-p)^n-1 < =1 maximum means equality occur only at n = 1 thus maximum is at n = 1 P(x=1) = p b. what is the probability that X is even? P(x=n) = (1-p)p + (1-p)^3 p +.................. since n is even now P(x=n at even) = p/(1-p) ( (1-p)^2 + (1-p)^4+..........) this is geometric distribution with ratio 1-p now P(x=n) = p/(1-p) *(1-p)^2/(1-(1-p)^2) = p/(1-p) *(1-p)^2(1/ 1 - p^2+2p-1) = p/(1-p) * (1- p)^2/2p-p^2) = p/(1-p) (1-p)^2/p(2-p) = (1-p)/(2-p) thus answer..

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- Let X be a geometric random variable with parameter p, and n and m be nonnegative integers. a. for what values of n is P(X=n) Maximum? b. what is the probability that X is even? Solution Let X be a geometric random variable with parameter p, and n and m be nonnegative integers. a. for what values of n is P(X=n) Maximum? P(x=n) = (1-p)^n-1 p we know that 0<=p<=1 now => 0<= 1- p <=1 (1-p)^n-1 < =1 maximum means equality occur only at n = 1 thus maximum is at n = 1 P(x=1) = p b. what is the probability that X is even? P(x=n) = (1-p)p + (1-p)^3 p +.................. since n is even now
- P(x=n at even) = p/(1-p) ( (1-p)^2 + (1-p)^4+..........) this is geometric distribution with ratio 1-p now P(x=n) = p/(1-p) *(1-p)^2/(1-(1-p)^2) = p/(1-p) *(1-p)^2(1/ 1 - p^2+2p-1) = p/(1-p) * (1- p)^2/2p-p^2) = p/(1-p) (1-p)^2/p(2-p) = (1-p)/(2-p) thus answer.