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 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.

Explanation:

Step1: Determine the number of non - zero digits

The non - zero digits are 1, 2, 3, 4, 5, 6, 7, 8, 9. So, to find the total number of outcomes for the 4 - digit lock code, 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 the number of non - zero even digits

The non - zero even digits are 2, 4, 6, 8. The total number of favorable outcomes is 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 $P(4,4)=\frac{4!}{(4 - 4)!}=\frac{4!}{0!}=4!=4\times3\times2\times1 = 24$.

Step3: Calculate the probability

The probability $P$ is the number of favorable outcomes divided by the total number of outcomes. So $P=\frac{24}{3024}=\frac{1}{126}$.

Answer:

The first blank is 9, the second blank is 3024, the third blank is 4, the fourth blank is 24.