Probability, geometric distribution

Question

What is the probability of $P(X<E(X))$ if $X$ has a geometric distribution such that $P(X=1)=0.25$.

I found $E(X)=0.25$. I guess I should find $P(X=0)$, but I don't know how. I can only plug $k>0$ values in the mass function.

Answer 1

We have that $$ \Pr\{X=k\}=(1-p)^{k-1}p $$ for $k\ge1$ and $$ \Pr\{X=1\}=(1-p)^0p=p=0.25 $$ so that $$ \operatorname EX=\frac1p=4. $$ We obtain that $$ \Pr\{X<\operatorname EX\}=\Pr\{X<4\}=\sum_{k=1}^3\Pr\{X=k\}=\sum_{k=1}^3(0.75)^{k-1}0.25\approx0.5781. $$