Asked by Taryn
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
Answered by
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
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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