mathprobability
posted by Tina .
A state lottery has a daily drawing to form a fourdigit number. The digits 1 through 9 are randomly selected for each of the four digits. For each selection any one of the digits 1 through 9 are possible. What is the probability to the nearest hundredth of the fourdigit number having at least one 3?

Use the "backdoor" approach
number of possible outcomes with no restriction = 9^4 = 6561
number of cases with "no 3" = 8^4 =4096
So the number of cases that have at least some 3's
= 65614096 =2465
prob(at least one 3) = 2465/6561 = .3757
or appr 0.38
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