Question

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?

Answers

Reiny
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

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