A state lotter has a daily drawing to form a four digit numb

A state lotter has a daily drawing to form a four digit number. The digits 1 through 9 are randomly selected for ach of the four digits. For each selection any one of the digits 1 through 9 are possible. What is the probability that the nearest hundreth of the four-digit number having at least one 3?

Solution

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 atleast some 3\'s = 6561-4096 = 2465

prob(atleast one 3) = 2465/6561 = 0.3537 or app. 0.38.