# PMF of a geometric random variable and binomial random variable

We have been discussing pmf. Can someone please show me

A. That the pmf for a binomial random variable sums to 1

B. That the pmf for a geometric random variable sum to 1

So I may see the difference

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What do you mean by A and 1 b? For any discrete random variable, the sum of the pmf over all possible values is $1$. This is because the total probability (i.e. the probability of the whole sample space) is always $1$. –  Robert Israel Sep 30 '12 at 23:59

A. You have to show $$\sum_{i=0}^n {n \choose i}p^i(1-p)^{n-i}=1$$ but the left hand side is $$(p+(1-p))^n=1.$$
B. Depending on your definition, you have to show $$\sum_{i=0}^\infty p(1-p)^{i}=1 \text{ or } \sum_{j=1}^\infty p(1-p)^{j-1}=1$$ and they are clearly the same sums if you replace $i$ by $j-1$.
The geometric series $\sum_{i=0}^\infty x^i = \frac{1}{1-x}$ for $|x|\lt 1$, so (taking the former) $$p\times \sum_{i=0}^\infty (1-p)^{i}=p\times \dfrac{1}{1-(1-p)}=1.$$