User IDs on an old computer system consist of 4 letter sequences from the first 8 letters of the English alphabet: a, b, c, d, e, f, g, h. Note that the same letter can be used any number of times in the 4 letter sequence. When a new user first registers, an ID is randomly generated(all outcomes equally likely) with the only condition that it is different from the IDs of all previously registered users.
a) How many different IDs are possible for the first registered user? (Just to control myself i found 8x8x8x8= 4,096)
b) What is the probability that the ID for the first registered user will start with the letter c and end with the letter g?
You correctly found that there are $8^4=4096$ possible equally likely ids that could be generated for the user.
Now... if it starts with a $c$ and ends with a $g$... then we have in essence $1$ option for the first character (it must be a c), $8$ options for the second character (it could be any of a,b,...,h), $8$ options again for the third character, and $1$ option for the final character (it must be a g).
Applying the rule of product, by multiplying the number of options available for each position we get the total number of passwords which begin with a $c$ and end with a $g$.
Dividing the result by the number of possible ids (noting that the ids are equally likely to occur) gives us the desired probability.
(Note: the number of "good occurrences" divided by the number of "possible occurrences" giving the probability is only a valid technique in the situation where each possible occurrence is explicitly known to be equally likely to occur. It may not be used in situations outside of that and can and will frequently lead to incorrect answers if this warning is ignored)