posted by Tina on .
A state lottery has a daily drawing to form a four-digit 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 four-digit number having at least one 3?
Use the "back-door" 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
= 6561-4096 =2465
prob(at least one 3) = 2465/6561 = .3757
or appr 0.38