# Assessing extreme probability parameter of Bernoulli events

Suppose we have an experiment which results in "success" with probability $$p$$ and in "failure" with probability $$1-p$$. The distribution of corresponding random value is assumed to be Bernoulli.

It turns out that after a hundred experiments there were no "successes". How can one assess $$p$$ with some confidence level (for example, $$0.95$$)?

My attempt. The probability that after a hundred experiments there will be no "successes" is $$\left(1-p\right)^{100}$$. So one can compare this probability with the given confidence level. That means that $$(1-p)^{100}\geq0.95$$ That gives us that $$0\leq p\leq 1-\sqrt[100]{0.95}\approx0.0005128$$ Is that approach correct? If no, how can one tackle that problem?

I don't think so. If you're trying to establish the range of $$p$$ over which the given result is not significant at the $$0.05$$ level, then you want
$$(1-p)^{100} \geq 0.05$$
and then $$0 \leq p \leq 0.0295$$ approximately. What you have there instead finds the range of $$p$$ over which the probability of obtaining a result that extreme is at least $$0.95$$, which I don't think is what you meant.
Alternatively, one can take a Bayesian approach, and use some appropriate prior (Beta distribution with $$\alpha = \beta = 1$$, which is uniform, will work), and update it according to the $$100$$ consecutive failures. The result (if you start with a uniform prior) is a Beta distribution with $$\alpha = 1, \beta = 101$$, whose PDF looks like this:
• I would argue that you should use $(1-p)^{100} \geq 0.025$ if you want to use a "two-sided" 95% interval where each tail contains a maximum mass of 0.025. This gives $[0, 0.0362167]$ as the confidence interval. – PhiNotPi Apr 23 at 17:50