Without looking at the labels, Adrien placed four CDs in four cases. What is the probability that exactly one of the CDs is in the wrong case? Express your answer as a common fraction.
+128566 This can't happen.....
If one CD is in the wrong case, then there must be (at least) two CDs in the the wrong case
For example.....if the correct CDs are in cases "3" and "4", but CD "1" is in case "2"....then CD "2" must be in case "1"...so....two CDs are in the wrong cases
