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.

The probability is found:

P = 24/3024 = 1/126

Step-by-step explanation:

To find total number of outcomes, we have to find permutation of 9 things taken 4 at a time:

P(9,4) = 9! / (9-3)!

P(9,4) =362880/120

P(9,4) = 3024

Number of even numbers from 0 to 9 = 4

To find desirable number of outcome, find permutation of 4 things taken 4 at a time.

P(4,4) = 4! / (4-4)!

P(4,4) = 24/1

P(4,4) = 24

The probability that the lock consists of all even digits is

P = No. of desirable outcomes / No. of total outcomes

P = 24/3024

step-by-step explanation:

the true statements are:

r = set of real numbers

n = set of natural numbers, 1,2,

w = n u{0} = {0,1,}

i cant answer this, theres no table. sorry man

step-by-step explanation: