a safe has a 4 - digit lock code that does not include zero as a digit and no digit is repeated. what is the…

Explanation:

Step1: Determine total number of non - zero digits

There are 9 non - zero digits (1, 2, 3, 4, 5, 6, 7, 8, 9). To find the total number of 4 - digit lock codes with no repeated digits, we find the permutation of 9 things taken 4 at a time. The formula for permutations is $P(n,r)=\frac{n!}{(n - r)!}$. Here, $n = 9$ and $r=4$, so $P(9,4)=\frac{9!}{(9 - 4)!}=\frac{9!}{5!}=9\times8\times7\times6=3024$.

Step2: Determine number of non - zero even digits

There are 4 non - zero even digits (2, 4, 6, 8). To find the number of 4 - digit lock codes with all non - zero even digits and no repeated digits, we find the permutation of 4 things taken 4 at a time. Using the permutation formula $P(n,r)$ with $n = 4$ and $r = 4$, we have $P(4,4)=\frac{4!}{(4 - 4)!}=\frac{4!}{0!}=4\times3\times2\times1 = 24$.

Step3: Calculate the probability

The probability $P$ is the number of favorable outcomes divided by the total number of outcomes. The number of favorable outcomes is 24 and the total number of outcomes is 3024. So the probability that the lock code consists of all even digits is $\frac{24}{3024}=\frac{1}{126}$.

Answer:

To find the total number of outcomes for this event, find the permutation of 9 things taken 4 at a time. The total number of outcomes is 3024. The total number of favorable outcomes is a permutation of 4 things taken 4 at a time. The probability that the lock code consists of all even digits is 24 out of 3,024.