a briefcase has a three - digit lock code that does not include zero as a digit. what is the probability…

Explanation:

Step1: Calculate total number of three - digit lock codes

Since the lock code is three - digit and does not include 0, for the first digit, there are 9 choices (1 - 9). For the second digit, there are 8 remaining choices (because one digit has been used), and for the third digit, there are 7 remaining choices. Using the permutation formula $P(n,r)=\frac{n!}{(n - r)!}$, here $n = 9$ and $r=3$, so the total number of three - digit lock codes is $P(9,3)=\frac{9!}{(9 - 3)!}=\frac{9!}{6!}=9\times8\times7 = 504$.

Step2: Calculate number of three - digit lock codes with all even non - zero digits

The non - zero even digits are 2, 4, 6, 8. So $n = 4$ and $r = 3$. Using the permutation formula $P(n,r)=\frac{n!}{(n - r)!}$, we have $P(4,3)=\frac{4!}{(4 - 3)!}=\frac{4!}{1!}=4\times3\times2=24$.

Step3: Calculate the probability

The probability is the number of favorable outcomes (lock codes with all even non - zero digits) divided by the number of total outcomes (all non - zero three - digit lock codes). So the probability is $\frac{24}{504}$.

Answer:

A. 24 out of 504