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 of a lock without all even digits? there 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 lock code without all even digits is 3000/3024.

Explanation:

Step1: Find number of all - even - digit lock codes

The non - zero even digits are 2, 4, 6, 8. The number of 4 - digit lock codes using these 4 non - zero even digits without repetition is given by the permutation formula \(P(n,r)=\frac{n!}{(n - r)!}\), where \(n = 4\) and \(r=4\). So \(P(4,4)=\frac{4!}{(4 - 4)!}=4!=24\).

Step2: Find number of non - all - even - digit lock codes

We know the total number of 4 - digit lock codes (without zero and no repetition) is 3024. The number of non - all - even - digit lock codes is \(3024-24 = 3000\).

Answer:

The probability of a lock code without all even digits is \(\frac{3000}{3024}\), and there are 24 ways to get a lock code with all even digits and 3000 ways to get a lock code without all even digits.