Aspire Faculty ID #17170 · Topic: AMU MCA 2020 · Just now
AMU MCA 2020

Suppose random variable $X$ has possible values $1,2,3,\dots$ and $P(X=j)=\frac{1}{2^j}$, $j=1,2,3,\dots$ then $P(X \text{ is even})$ equals

Solution

Even values: $2,4,6,\dots$

$P(\text{even})=\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+\dots$

$=\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\dots$

This is geometric series:

$=\frac{1/4}{1-1/4}=\frac{1/4}{3/4}=\frac{1}{3}$

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