Consider a probability distribution on the numbers $0, 1, 2, …, 20$ with probabilities given $P(k) = \frac a{2^k}$ .

Question

Find $a$ so that the sum of all probabilities is $1$, and hence $P(k)$ is a well-defined formula for a probability distribution.

Answer 1

If $P(k) = \frac{a}{2^k}$ for $0\leq k \leq 20$, then the fact that $\sum_{k=0}^{20} P(k)=1$ implies that $$ 1 = a\sum_{k=0}^{20} \frac{1}{2^k} = a\frac{1-\frac{1}{2^{21}}}{1-\frac{1}{2}} = 2a\left( 1-\frac{1}{2^{21}}\right) $$ i.e. $$ a = \frac{1}{2\left( 1-\frac{1}{2^{21}}\right)}. $$

You can check that this necessary condition is sufficient for $P$ to be a probability distribution on $\{0,\dots,20\}$: it is a non-negative function taking values in $[0,1]$ and summing to $1$.