They need to have common denomenators so change the fractions to 12/15 - 5/15 which is 7/15.
For the first digit there are 7 choices. For the second digit there are 6 choices (because we can't use the same one again). And so on, until there is only one choice for the last digit.
The number of possible codes is this:
7*6*5*4*3*2*1 = 7! = 5040
For any distribution, the sum of the probabilities of all possible outcomes must be 1. In this case, we have to have

We're told that
, and we're given other probabilities, so we have

The expected number of calls would be
![E[X]=\displaystyle\sum_xx\,P(X=x)](https://tex.z-dn.net/?f=E%5BX%5D%3D%5Cdisplaystyle%5Csum_xx%5C%2CP%28X%3Dx%29)
![E[X]=0\,P(X=0)+1\,P(X=1)+\cdots+4\,P(X=4)](https://tex.z-dn.net/?f=E%5BX%5D%3D0%5C%2CP%28X%3D0%29%2B1%5C%2CP%28X%3D1%29%2B%5Ccdots%2B4%5C%2CP%28X%3D4%29)
![E[X]=1.4](https://tex.z-dn.net/?f=E%5BX%5D%3D1.4)