A safe has a 4-digit lock code that does not include zero as a digit and no digit is repeated. What is the probability that the lock code consists of all even digits?

Answer 1

To solve for probability of an event, the simplest way is using:

(Number of outcomes of given event occurring)/(Total number of outcomes)

There are five even single digits:

0, 2, 4, 6, 8

0 is not a possible digit as per the givens. This leaves you with:

2, 4, 6, 8

No digit may repeat, and there are 4 digits. This means a factorial can be used to find how many possible combinations fit the criteria:

4! = 24 = Number of outcomes that fit the criteria

A factorial, being the number multiplied by every previous number down to 1, is used because if you have 4 blank spaces with no repeats, the first space has 4 possibilities. Given the lack of repetition, the second space can only be one of 3 digits with the first space given. The third space can only be 2 digits, and the last space can only be one. Therefore, you get 4*3*2*1, which equals 24.

For the number of total outcomes, use a similar strategy. Part of There are four spaces. The first can be any of 9 digits. The second can be 8, the third 7, and the fourth 6. Therefore, you get 9*8*7*6, which equals 3024.

Therefore the probability is 24/3024, or .000793650794