Every box of cereal contains one toy from a group of five distinct toys, each of which is mutually independent from the others and is equally likely to be within a given box. On average, how many boxes of cereal will you need to open in order to collect all five toys?
i'm doing 5/5 + 5/4 + 5/3 + 5/2 + 5/1 = 137/12 is this correct?
Yes, what you're doing is correct. You have five different toys with the same probability of being in a box and mutually independent from the others.
To get the first toy, you only need to open 1 box, whatever toy is in there works (that's your 5/5).
To get the second toy, you now have 4 toys left, and each box has a probability of 4/5 to contain a toy you don't have. So on average you need to open 1/(4/5) or 5/4 boxes to get the second toy.
To get the third toy, now each box has a 3/5 probability of containing a toy you don't have, meaning on average you need to open 1/(3/5) or 5/3 boxes to get the third one.
To get the fourth toy, each box has a 2/5 probability of containing a toy you don't have, and you need to open, on average, 1/(2/5) or 5/2 boxes to get it.
Finally, to get the fifth toy, each box has a 1/5 probability of containing a toy you don't have, and you need to open, on average, 1/(1/5) or 5/1 boxes to get it.
So, the total is, as you have correctly identified:
$$5/5 + 5/4 + 5/3 + 5/2 + 5/1 = 137/12$$
