You need more information, i.e. the probability of getting a cd
and each of the CDs have equal probability of being placed in a cereal box,
then you can find the probability of NOT finding all five CDS in the 12 boxes
by calculating the probabilities that
(1) one of the CDs is not in the 12 packets
(2) two of the CDs are not in the 12 packets
(3) three of the CDs are not in the 12 packets
(4) four of the CDs are not in the 12 packets
All 5 CDs cannot be missing of course.
The probability that one CD is missing is 5C1(4/5)^(12)
because one of the other 4 CDs are in each box for each missing one.
The probability that two CDs are missing is 5C2(3/5)^(12)
because one of the other 3 CDs is in each box for every missing pair.
The probability that three CDs are missing is 5C3(2/5)^(12)
Also, include four CDs missing.
Now the question is...
Does the probability of one being missing include the probability of 2, 3 or 4 also missing ?
Subtract the probability of at least one CD missing from 1.