The following question has me extremely confused:
At a certain university, registrations for courses have to be made over the telephone. There are so many calls that $90$% of the time you will get a busy signal. When you get a busy signal, you hang up and dial again.
Suppose that you have gotten a busy signal the last $3$ times you have called.
Calculate the expected value of the additional number of unsuccessful tries before you get through for the first time.
I was working under the assumption that the expected value of a geometric distribution is $1$ over the probability of success so in this case, $10$. However the solution for this problem uses the fact this distribution is memoryless and so the expected value is $0.9$ over $0.1=9$. Why is this definition of expected value being used as opposed to the one I'm familiar with?
The geometric law is memoryless thus $P(X=k)=P(X=k+h|X>h)=P(X=k)$
this means that (as known) $E(X)=\frac{1-p}{p}=9$ is the same as the expected value of the additional number of unsuccessful tries before you get through for the first time, and this is valid for any numbers of consecutive insuccesses.
This because there are two parametrization of geometric distribution, one counting the number of trial before the first success and another one counting the failures before the first success. This case is this second one. The relationship between the two distribution is the following
$$X=Y-1$$
thus the expectation, using the formula you are familiar with will be
$$E(X)=\frac{1}{0.1}-1=10-1=9$$