Asafe 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
ings 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.

the probability is of 0.79%

Step-by-step explanation:

There are 4 numbers, 0 not included, and there is no number repeated.

First, the total number of combinations will be:

c1 = 9*8*7*6 = 3024

this is because the first number has 9 possibilities to chose, the second will have 8 possibilities, because one number is already taken, and so on.

Now, we have 4 even numbers: 2, 4, 6, 8

the total possible combinations of those 4 numbers are calculated in the same way: the first number has 4 possibilities, the second has 3, and so on

c2 = 4*3*2*! = 24

now, the probability that the code consists only in even numbers is equal to the number of combinations of only pair numbers divided the total number of combinations:

p = c2/c1 = 24/3024 = 0.0079

for a percentage, we multiply this number by 100%

0.0079*100% = 0.79%

55? i think maybe

step-by-step explanation:

they are simplified

step-by-step explanation: