Answer
(a) $P(X=2)=3/64=0.046875$
(b) $P(X≤2)=63/64=0.984375$
(c) $P(X>2)=1/64=0.015625$
(d) $P(X≥1)=0.25$
Work Step by Step
Given that,
$f(x)=(3/4)(1/4)^{x}$, then,
$f(0)=(3/4)(1/4)^{0}=3/4=0.75$,
$f(1)=(3/4)(1/4)^{1}=3/16=0.1875$,
$f(2)=(3/4)(1/4)^{2}=3/64=0.046875$,
......., we can find that for all possible values of x,
f(x) ≥ 0,
and, $P(X=0)+P(X=1)+P(X=2)+......=(3/4)(1/4)^{0}+(3/4)(1/4)^{1}+(3/4)(1/4)^{2}+...=(3/4)[1+(1/4)^{1}+(1/4)^{2}+...]=(3/4)[1/(1-(1/4))=1$
(a) $P(X=2)=3/64=0.046875$
(b) $P(X≤2)=P(X=0)+P(X=1)+P(X=2)=3/4+3/16+3/64=63/64=0.984375$
(c) $P(X>2)=1-P(X≤2)=1-63/64=1/64=0.015625$
(d) $P(X≥1)=1-P(X<1)=1-P(X=0)=1-0.75=0.25$
