Question

Suppose that X is a geometric random variable with p=2/3 Solve for P (X=1 ∩ X ≥ 3)?

Answer 1

$X$ can't simultaneously be equal to 1 and larger than 2, so the probability must be zero.

Did you mean to find the probability of the union? In that case, since the events are mutually exclusive, you would be able to write

$\mathbb P((X=1)\cap(X\ge3))=\mathbb P(X=1)+\mathbb P(X\ge3)$
$=\mathbb P(X=1)+1-\mathbb P(X$
$=\mathbb P(X=1)+1-\mathbb P(X\le2)$
$=1+f_X(1)+F_X(2)$

where $f_X(x)$ is the probability density function and $F_X(x)$ is the cumulative distribution function for the random variable $X$.