The complete question is;
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?
To find the total number of outcomes for this event, find the permutation of things taken 4 at a time. The total number of outcomes is: The total number of favorable outcomes is a permutation of things taken 4 at a time. The probability that the lock code consists of all even digits is out of 3,024
Answer:
A) The total number of outcomes = 3024
B) The probability that the lock code consists of all even digits is 24 out of 3024
Step-by-step explanation:
To find the total number of outcomes for this event, since it doesn't include zero as a digit, thus there are 9 possible digits. So let's find the permutation of 9 things taken 4 at a time.
Thus;
The total number of outcomes = P(9,4) = 9!/(9 - 4)! = 9!/5! = 3,024
The total number of favorable outcomes is a permutation of 4 things taken 4 at a time.
Thus;
The probability that the lock code consists of all even digits =
P(4,4) = 4!/(4 - 4)! = 4!/0! = 24/1 = 24
That's 24 out of 3,024.
