# Probability of number of events where $p_{n+1} = \frac{1}{5}p_n$

For $n \ge 0, p_{n+1} = \frac{1}{5}p_n$. Where $p_n$ is the probability of an envent occuring n times in a period.

What is the probability that more than one event occurs?

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$1= \sum\limits_{n=0}^{\infty}(\frac{1}{5})^{n}p_0$

$(\frac{1}{5})(1) = (\frac{1}{5})\sum\limits_{n=0}^{\infty}(\frac{1}{5})^{n}p_0$
$1 - (\frac{1}{5})(1) = p_0 (\frac{1}{5})^{0}$
$\frac{4}{5} = p_0$

using the formula in the question,
$p_1 = \frac{1}{5}(\frac{4}{5})$
$p_1 = \frac{4}{25}$

$P(n\ge 2) = 1 - P(n=0) - P(n=1)$
$P(n\ge 2) = 1 - \frac{4}{5} - \frac{4}{25}$
$P(n\ge 2) = 0.04$

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The probability of more than one event is $1-p_0-p_1$, which is $1-p_0-\frac{p_0}{5}$.

Note that $p_k=p_0\left(\frac{1}{5}\right)^k$. But $p_0+p_1+p_2+\cdots=1$. Thus $$p_0\left(1+\frac{1}{5}+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+ \cdots\right)=1.$$ Summing the geometric series, we find that $p_0(5/4)=1$.

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