Finding values of a constant in a probability distribution

Question

A probability distribution for the random variable $X$ is defined by: $$\mathbb{P}[X=x] = K\cdot(0.9)^x,\quad x = 0,1,2,\ldots$$ It is asked to find $\mathbb{P}[X\geq 2]$.

When there is a domain for $x$, that is $[0,n]$ I am able to find the value of $K$ by applying the rule that sum of probabilities add up to $1$. But in this case I am not given a domain for $x$...

$$\mathbb{P}[X\geq 2] = 1 - \left(\mathbb{P}[X=0]+\mathbb{P}[X=1]\right) = 1 - 1.9K,$$

so the answer is $0.81$... how do you get this?

Answer 1

The sum of probabilities must still be one:

$$1=\sum_{n=0}^{\infty} P(X=n)=\sum_{n=0}^{\infty} k \left(\frac9{10}\right)^n=\frac{k}{1-\frac9{10}}=10k$$

Hence $k=\frac{1}{10}$.

Answer 2

Answer:

The domain is essentially, 0 to infinity.

$$\sum_{0}^{\infty}k(0.9^x) = k{(1+0.9+.0^2+.9^3+\cdots)} = k(\frac{1}{(1-.9)})=1 $$

$$ k = 0.1$$

$$P(X>=2) =1 - 0.1*0.9^0+0.1*0.9^1 = 1-.1-.09 = 0.81$$