A person has forgotten the PIN code for his SIM card, but remembers that it contains three "fives", and one of the numbers is either a "seven" or "eight". What is the probability of successful login on the first try?
So, i think that because there are two case one of the numbers is either a "seven" or "eight", the answer must be a sum of probabilities in cases of seven and eight. The number of permutations with repetitions in case of 7 is $\frac{4!}{3!1!}=4$, so probability is $\frac{1}{4}$. The same in case of eight: $\frac{1}{4}$. So the answer is $\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$. However book says that the correct answer is $\frac{1}{8}$. I see, if i write all possible combinations, but why does my formula doesn't work? Isn't it "or" case?
If there are two cases (7 or 8) you need to sum them in the sample space, so you get $\tfrac{1}{4+4}$. You sum probabilities when there are several possibilities to get to the desired outcome, however here the outcome (the correct code) is unique.