What is the probability that a random 6-digit number will have at least one 0, at least one 1, and at least one 2? Hi this a question from my textbook: A first course in probability, It doesn't have the solution so I'm curious as to what the answer is. This is the question:
What is the probability that a random 6-digit number will have at least one 0, at
least one 1, and at least one 2? (the number can't start with a 0).
$My Answer$
Well is know that all the possible combinations of a 6 - digit number that doesn't start with $0$ is: $9(10^5)$
Next I considered the all the possible combinations of a $6$ - digit number not having at least one 0, at least one 1, and at least one 2 which is $7^6$
so: $1 - 7^6/9(10^5) = 0.87 $
is this correct?

Hint:
event A - number has at least one 0;
event B - number has at least one 1;
event C - number has at least one 2;
A and B and C = not[ (not A) or (not B) or (not C) ]
Prob[ (not A) or (not B) or (not C) ] = Prob(not A) + Prob(not B)+ Prob(not C) - Prob[(not A) and (not B)] - Prob[(not A) and (not C)] - Prob[(not C) and (not B)] + Prob[(not A) and (not B) and (not C)]
