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

a briefcase has a three - digit lock code that does not include zero as a digit. what is the probability that the lock code consists of all even digits if the same digit is not used more than once in the lock code? 24 out of 504 60 out of 504 24 out of 84 60 out of 84

a briefcase has a three - digit lock code that does not include zero as a digit. what is the probability that the lock code consists of all even digits if the same digit is not used more than once in the lock code? 24 out of 504 60 out of 504 24 out of 84 60 out of 84

Answer

Explanation:

Step1: Calculate total number of 3 - digit codes

There are 9 non - zero digits (1 - 9). The number of ways to form a 3 - digit code with no digit repeated is given by the permutation formula $P(n,r)=\frac{n!}{(n - r)!}$, where $n = 9$ and $r=3$. So, $P(9,3)=\frac{9!}{(9 - 3)!}=\frac{9!}{6!}=9\times8\times7 = 504$.

Step2: Calculate number of 3 - digit codes with all even digits

There are 4 non - zero even digits (2, 4, 6, 8). The number of ways to form a 3 - digit code with these 4 digits and no digit repeated is $P(4,3)=\frac{4!}{(4 - 3)!}=\frac{4!}{1!}=4\times3\times2=24$.

Step3: Calculate the probability

The probability $P$ is the number of favorable outcomes divided by the number of total outcomes. So, $P=\frac{24}{504}$.

Answer:

24 out of 504