Let k >= 3 be any integer. What is the probability that a random k-digit number will have at least one 0, one 1 and one 2? (as usual every number starts with either 1,2,...9 and NOT zero)
b(x : n,p) = (n x)p^x*(1-p)^(n-x) where x = 0, 1, 2, ... ,n
This equation represents a binomial probability distribution. The outcome is either a success or a failure. In this case, a success would be a number containing at least a 0, 1 or 2. A failure would be otherwise.
The Attempt at a Solution
I'm not sure how to apply this equation in this situation, especially with the constraint put on k. I know that the number of "trials" must be 1. I also know that the outcome is either a success or a failure. I'm assuming that I'm looking for p in the above equation, which is the probability of success. Perhaps I'm missing something? I'm not sure, I've been attempting this for a while now with no luck.
Any advice to point me in the right direction would be great. Thanks