a number between 1000 and 9999, inclusive, is chosen at random. what is the probability that it contains a)no 9's? b) at least one 9?

Answer 1

There are 9000 numbers from 1000 to 9999 inclusive

then number without any 9 are
8x9x9x9 = 5832

reasoning:
the lead digit has to be
1,2,3,4,5,6,7,8
after that each place can be filled in 9 ways, that is, with
0,1,2,3,4,5,6,7,8

prob (no 9 anywhere) = 5832/9000 = .648

so prob(at least one 9) = 1 - .648 = .352