The number of ways the serial number can contain none of the three numbers, or be composed of only digits 0 to 6 is 78.
The number of ways the serial number can contain at least one 7 but no 8 or 9 is 88-78. We must subtract 78 because there are that many numbers that contain only the digits 0 to 6 but no 7. Likewise the number of ways to add at least one 8 or one 9 is also 88-78. So the number of ways the serial numbers will contain just one digit out of {7,8,9} is 3*(88-78).
Now let's consider the number of ways the serial number can have at least one 7 and 8 but no 9. The number of all combinations leaving out the 9 is 98. However some of those have no 7 or 8 digits. The number of combinations with 7 but no 8 is 88-78. This is also the number of combinations with 8 but no 7. We must also subtract the number of combinations with only 0-6, which is 78. So the number of combinations with at least one 7 and one 8, but no 9, is 98-2*(88-78)-78. Mutliply this by three for the number of ways the serial number can omit the 7 and 8 but leave the other two chosen numbers. So the probability the serial number will have 2 out of our 3 numbers is 3*(98-2*(88-78)-78).
So the probability that the serial number will not contain all three chosen numbers is:
(78 +
3*(88-78) +
3*(98-2*(88-78)-78)))/108 =
(3*98-3*88+78)/108 = 0.84573316.
To get the probability will contain all three chosen numbers just subtract the number above from one, or 1 - (3*98-3*88+78)/108 = 0.15426684.
Michael Shackleford, A.S.A.