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
To find the probability of a four-digit number having at least one 3, we first need to determine the total number of possible four-digit numbers and the number of four-digit numbers without any 3.
Step 1: Calculate the total number of possible four-digit numbers
Since each digit can be any number from 1 to 9, there are 9 possibilities for each digit. Thus, the total number of possible four-digit numbers is 9 × 9 × 9 × 9 = 6561.
Step 2: Calculate the number of four-digit numbers without any 3
In this case, we need to consider that each digit can be any number from 1 to 9 except 3, so for each digit, there are 8 possibilities. Therefore, the number of four-digit numbers without any 3 is 8 × 8 × 8 × 8 = 4096.
Step 3: Calculate the number of four-digit numbers with at least one 3
To find this, we subtract the number of four-digit numbers without any 3 from the total number of possible four-digit numbers:
6561 - 4096 = 2465.
Step 4: Calculate the probability
The probability is given by the number of favorable outcomes (four-digit numbers with at least one 3) divided by the total number of possible outcomes (all four-digit numbers):
P = favorable outcomes / total outcomes
P = 2465 / 6561
P ≈ 0.3757
So, the probability to the nearest hundredth of the four-digit number having at least one 3 is approximately 0.38.