Four fair dice D1, D2, D3 and D4, each having six faces numbered $1$ through $6$, are rolled simultaneously. What is the probability that D4 shows a number appearing on one of D1, D2, or D3?
My approach:
Case 1:
Let all the three dice give the same output:
$$\frac{6}{6}\text{ (any number comes on first)}\times\frac{1}{6}\times\frac{1}{6}\times\frac{1}{6}.$$
Case 2:
Let two be the same:
$$^3C_1\times\frac{6}{6}\times{}^2C_1\times\frac{1}{6}\times\frac{5}{6}\times{}^2C_1\times\frac{1}{6}.$$
Case 3:
Let all three be different:
$$\frac{6}{6}\times\frac{5}{6}\times\frac{4}{6}\times{}^3C_1\times\frac{1}{6}.$$
Adding, we get $\frac{91}{216}$, which is the correct answer. Is my method correct, and is there any other simpler method?
Hint for an easier solution:
The event that $D_4$ shows a number that is not on $D_1,D_2,D_3$ can also be described as the event that $D_1,D_2,D_3$ show a number that differs from the number on $D_4$.