Four fair dice D1, D2, D3 and D4, each having six faces numbered 1, 2, 3, 4, 5 and 6, are rolled simultaneously. The probability that D4 shows a number appearing on one of D1, D2 and D3 is

Four fair dice rolled simultaneously.
Total number of outcomes = 6 × 6 × 6 × 6 = 1296
Suppose, the outcome of the last one does not appear in the first three.
The number of such outcomes possible = 6 × 5 × 5 × 5 (D1 can show any of 6 numbers and the first three can show any of the remaining 5 numbers)
Hence, probability that D4 does not show a number shown by any of D1, D2 or D3 = (6 × 5 × 5 × 5) / 6^4 = 750/1296
So, the required probability = 1 - 750/1296 = 546/1296 = 91/216