# Finding the Probability of a random variable with countably infinite values

So I was working on a problem where I am provided with a PMF $p_X(k)= c/3^k$ for $k=1,2,3....$ I was able to calculate $c$ using the basic property of PMF and it came to be 2. I am not able to solve the next part which states that "Find $P(X\ge k)$ for all $k=1,2,3......$.

Any suggestions?

P.S :Here is the actual question:

Let X be a discrete random variable with probability mass function $p_X(k) = c/3^k$ for k = 1, 2, ... for some $c > 0$. Find $c$. Find $P(X\ge k)$ for all $k = 1, 2,3....$

## 1 Answer

Consider, $$S=\frac{2}{3}+\cdots+\frac{2}{3^{k-2}}+\frac{2}{3^{k-1}}$$

multiply $S$ by $\frac{1}{3}.$ Thus,

$$\frac{1}{3}S=\frac{2}{3^{2}}+\cdots+\frac{2}{3^{k-1}}+\frac{2}{3^{k}}.$$

Subtract $S$ of $\frac{1}{3}S,$

$$\frac{2}{3}S=\frac{2}{3}-\frac{2}{3^{k}}.$$ Thus,

$$S=1-\frac{1}{3^{k-1}}.$$ Now,

if $k=1, P(X\geq k)=1,$ and if $k>1,$

\begin{eqnarray} P(X\geq k)&=&1-P(X< k)\\ &=&1-P(X\leq k-1)\\ &=&1-\displaystyle \sum_{n=1}^{k-1}\frac{2}{3^{n}}\\ &=&1-(1-\frac{1}{3^{k-1}})\\ &=&\frac{1}{3^{k-1}} \end{eqnarray}