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Suppose I flip a coin until I get a tails. Let X be the number of flips this takes. What is the probability that there are at least 4 days in a year where I needed more than 5 flips to get a tails?

Is this kind of question combining geometric and exponential distributions with some kind of conditional probability? I'm unsure how to compute this.

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(making a few silly but necessary assumptions such as the flips occur fast enough that you will always finish flipping by the end of the day, and that you forgot to mention that flipping occurs each day of the year, that the coin is fair, and that events are independent)

Let $A_i$ be an indicator random variable: $$A_i=\begin{cases} 1&\text{if it takes more than}~5~\text{flips to get a tails on day}~i\\ 0&\text{otherwise}\end{cases}$$

Let $A = \sum\limits_{i=1}^{365} A_i$

By definition, $A$ is binomially distributed with $n=365$ trials and $p=Pr(A_i=1)$

What is $Pr(A_i=1)$? To see this, notice that one of exactly two situations will occur: The first tail is flipped within the first five attempts, or the first five attempts are all heads.

We see then that it doesn't actually matter to this problem when the first tail is, just what the outcomes of the first five flips are. We see that $Pr(A_i=1)=Pr(\text{five out of five flips all come up heads})=(\frac{1}{2})^5 = \frac{1}{32}$

The problem asks us to calculate $Pr(A\geq 4)$.

Remembering that $Pr(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$ for a binomially distributed random variable, and applying the rule of complementary events, we have:

$Pr(A\geq 4) = 1 - Pr(A=0)-Pr(A=1)-Pr(A=2)-Pr(A=3) \\ = 1-\binom{365}{0}\frac{31^{365}}{32^{365}}-\binom{365}{1}\frac{31^{364}}{32^{365}}-\binom{365}{2}\frac{31^{363}}{32^{365}}-\binom{365}{3}\frac{31^{362}}{32^{365}}$

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