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 number of ways to get a lock code without all even digits? there are to get a lock code with all even digits. there are 3,024 different 4 - digit lock codes. there are to get a lock code without all even digits. the probability of a 4 - digit lock code without all even digits is

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 $n = 4$ (the number of non - zero even digits) and $r=4$ (the number of digits in the code). So $P(4,4)=\frac{4!}{(4 - 4)!}=4!=4\times3\times2\times1 = 24$.

Step2: Calculate number of lock codes without all even digits

We know the total number of 4 - digit non - zero non - repeated lock codes is 3024. The number of lock codes without all even digits is the total number of lock codes minus the number of all - even - digit lock codes. So it is $3024-24 = 3000$.

Step3: Calculate the probability

The probability $P$ of getting a lock code without all even digits is the number of lock codes without all even digits divided by the total number of lock codes. So $P=\frac{3000}{3024}=\frac{125}{126}$.

Answer:

There are 24 ways to get a lock code with all even digits.
There are 3000 ways to get a lock code without all even digits.
The probability of a 4 - digit lock code without all even digits is $\frac{125}{126}$.