Number of ways the first 4 databases can be selected = 9C4 = 126 Probability that it is in 2 of the first 4 = 5C2 / 126 = 10/126 Probability that it is in 3 of the first 4 = 5C3 / 126 = 10/126 Probability that it is in 4 of the first 4 = 5C4 / 126 = 5/126
Probability that it is in at least 2 of the first 4 = 10/126 + 10/126 + 5/126 = 25/126