Question

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 of a lock without all even digits? there are 3,024 different 4 - digit lock codes. the probability of a 4 - digit lock code without all even digits is 1/3024, 24/3024, 3000/3024, 3023/3024

Explanation:

Step1: Calculate number of all - even - digit lock codes

The non - zero even digits are 2, 4, 6, 8. The number of ways to arrange 4 non - repeated non - zero even digits is given by the permutation formula $P(n,r)=\frac{n!}{(n - r)!}$, where $n = 4$ and $r=4$. So $P(4,4)=\frac{4!}{(4 - 4)!}=4!=24$.

Step2: Calculate probability of non - all - even - digit lock codes

The total number of 4 - digit lock codes with no zero and no repeated digits is 3024. The number of 4 - digit lock codes with all even digits is 24. The number of 4 - digit lock codes without all even digits is $3024−24 = 3000$. The probability $P$ of a 4 - digit lock code without all even digits is $\frac{3000}{3024}$.

Answer:

$\frac{3000}{3024}$