Of all the four-digit positive integers containing only digits from the set {2,4,6,8} what fraction of them have at least one 8?
The total number of 4 digit numbers with digits from the set given is \((4)(4)(4)(4) ={4}^{4}.\) If we leave out 8, the total number changes to
\((3)(3)(3)(3)={3}^{4}\). So the fraction having at least one 8, which is 1 minus the fraction having no 8, would then be
\(1-\frac{{3}^{4}}{{4}^{4}}=1-\frac{81}{256}=\frac{175}{256}\)
