Question

Four numbers from the set 0 to 9, including 0 and 9, are used to create a personal identification number. Once a number is used, it cannot be reused.

How many possible outcomes exist for the situation?


How many possible outcomes begin with a 0?


What is the theoretical probability of choosing a personal identification number that begins with a 0?

Answer 1

Total number of 4-digit PINs (distinct digits, must include 0 and 9):

- Choose the other 2 digits from the remaining 8: C(8,2) = 28.
- Arrange the 4 chosen digits: 4! = 24.
Total = 28 × 24 = 672.

Number that begin with 0:

- Fix first digit = 0. Choose 2 other digits from the remaining 8: C(8,2) = 28.
- Arrange the remaining 3 digits: 3! = 6.
Total = 28 × 6 = 168.

Theoretical probability a randomly chosen such PIN begins with 0:
168 / 672 = 1/4 = 0.25 = 25%.

(Also intuitively, 0 is equally likely to occupy any of the 4 positions, so probability = 1/4.)