Assume that when a man calls a restaurant to order some food, his call has a probability $0.2$ of reaching through. Every attempt is independent and we can assume that the restaurant is open every day.
What is the probability that over the course of $30$ days, the man on average has to call more than $3$ times (call going through on the last one), every day?
I initially tried setting this up as a binomial distribution but with no luck, as I don't see how to deal with the average in this case.
Model the number of calls until, and including, the first success as a Geometric RV and the number of days where one has to call more than 3 times to get through as Binomial as follows:
Each day the man calls the restaurant until they answer, which occurs with probability $p=0.2$, each call being independent of one another. Let $W$ be the number of the call that is finally answered. Then $W\sim Geo(p,\mathbb{N})$, that is $P(W=k)=q^{k-1}p$, where $q=1-p=0.8$. So, for example, the probability they answer on the first call is $P(W=1)=p=0.2$, and getting through in at most $k$ calls is given by the CDF, $P(W \leq k )=1-q^k$. Therefore, $P(W>3)=q^3$.
Now, if we repeat this for 30 days, assuming the calls for each day are independent of the previous days, we can model the number of days where it takes more than 3 calls to get through as a binomial RV, $Y\sim B(n,p^*)$, where $n=30$ and $p^*=P(W>3)$. You then are seeking to calculate $P(Y=30)$, the probability that every day the man must call more than 3 times to get through, which comes out to $P(Y=30)=1.89713759\cdot 10^{-9}$
The number of days the man must call more than 3 times to get through, on average is given by $E(Y)=np^*=30\cdot(0.8)^3=15.36$