Recently I realized that lost all computing skill in probability. Please take a look at the following problem.
There are $b$ balls that are thrown one by one and bounce from top to bottom on the $n$ steps of stairs. The balls are thrown one by one with one unit of time interval when the previous ball touches the first step the next ball is thrown. Reaching the next step takes the same one unit of time. The probability that the ball miss the step is $p$. The goal is achieved when at least one ball touches the last step.
What is the maximum running time of the process till the success in the worst case?
I think the worst case is when we lost first $b-1$ balls on the first step, all $b-1$ balls missed the first step and the last $b$th last successfully reached the last step, therefore the worst running time is $b-1+n$.
What is the probability of a success (at least one ball reached the last step)?
$P(success)=1-p^b$
What is the smallest number of balls should be thrown to guarante that the probability of success is at least $1 - \frac{1}{n}$.
$$1-p^b \geq 1 - \frac{1}{n}$$
$$p^b \leq \frac{1}{n}$$
$$b \cdot \log p \leq - \log n$$
$$b \leq - \frac{\log n}{\log p}$$
I am not sure if I get it correct.
Give an upper bound on the expected running time.
Let try to use a Geometric distribution
$$E=\sum_{i=1}^{b-1} i \cdot p^i \cdot (1-p)^n$$
I still not sure if I get it right with expected running time
Answer to part 1 is spot on.
Answer to part 2, the trick is to realise that these are independent events- each ball hits the last step with probability $p$.
So the chance that at least 1 ball hits is the opposite that none hit i.e. $1-(1-p)^b$
Use this for Part 3 and post what you get.