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A random variable X takes the values `0,1,2,3,...,` with prbability `PX(=x)=k(x+1)((1)/(5))^x`, where k is a constant, then `P(X=0)` is.A. `7/(25)`B. `(18)/(25)`C. `(13)/(25)`D. `(16)/(25)` |
Answer» Correct Answer - D `P(X=0)=k,P(X=1)=2k(1/5)^1` `P(X=2)=3k(1/5)^2...=1` Since `P(X=0)+P(X=1)+P(X=2)+...=1` `thereforek+2k(1/5)+3k(1/5)^2+...=1` `and (k/5+2k(1/5)^2+...=1/5)/(k+k(1/5)+k(1/5)^2+...=4/5)` `rArr" "k/(1-1/5)=4/5rArrk=(16)/(25)` `therefore" "P(X=0)=(16)/(25)(0+1)(1/5)^0=(16)/(25)` |
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