I am practising for an exam that has a section on introductory probability theory, we have covered basic Markov chains and conditional probability + binomial distribution, not much more. In these first three questions, we encounter probability, the remainder of the question focuses on modelling this as a Markov chain. I am specifically already having problems with the first question, which doesn't allow me to move on.
During his daily walk a donkey encounters $n=10$ stones. In parts a), b) and c) assume that the probability of the donkey bumping into the same stone twice is equal to $p=\frac 1 3$, independent of all other events. For all these three parts give a clear explanation and computation.
Question a: What is the probability of the donkey bumping into exactly one stone during a walk? Express this probability in terms of $n$ and $p$, you do not need to give a numerical answer.
Question b: What is the expected number of times the donkey bumps into a stone during a walk?
Question c: What is the probability of the donkey bumping into the last stone, given that he has already bumped into 5 of the other 9 stones?
I am really not sure what is expected of me, I am kind of confused by the fact that they only give me the probability for hitting the same stone twice, but don't really tell me how probable it is that a donkey hits his head on a stone. My best guess would be $\frac{1}{\sqrt{3}}$. Part $c$ would go easily if I choose some clever random variable and use $P(A \cap B)= P(A|B)P(B)$. Can I get some pointers? I have been stuck on this problem even though it seems so simple.
a) Since the events are independent, then probability bumping in to a stone once is $\sqrt{p}=q$ so bumping once in one of $10$ stones is $$P= {10\choose 1}q(1-q)^9$$
b) If $X$ counts the number of bumpings, and $X_i$ is indicator variable for bumping in to $i-$th stone, then $$E(X) = \sum _{i=1}^{10}E(X_i) = \sum _{i=1}^{10}P(X_i=1) =10\sqrt{p}$$
c) Since the events are independent it is $\sqrt{p}$.