theoretical and experimental probability\ninstruction active\nanalyzing complements\na safe has a 4 - digit…

theoretical and experimental probability\ninstruction active\nanalyzing complements\na 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?\nthere \nthere are 3,024 different 4 - digit lock codes.\nthere are \nto get a lock code without all even digits.\nthe probability of a 4 - digit lock code without all even digits is

theoretical and experimental probability\ninstruction active\nanalyzing complements\na 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?\nthere \nthere are 3,024 different 4 - digit lock codes.\nthere are \nto get a lock code without all even digits.\nthe probability of a 4 - digit lock code without all even digits is

Answer

Explanation:

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

The non - zero even digits are 2, 4, 6, 8. The number of permutations of 4 non - repeated non - zero even digits 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: Calculate number of lock codes without all even digits

The total number of 4 - digit lock codes with no zero and no repeated digits is 3024. The number of lock codes without all even digits is $3024-24 = 3000$.

Step3: Calculate the probability

The probability $P$ of a 4 - digit lock code without all even digits is $P=\frac{3000}{3024}=\frac{125}{126}$.

Answer:

There are 24 ways to get a lock code with all even digits. There are 3000 ways to get a lock code without all even digits. The probability of a 4 - digit lock code without all even digits is $\frac{125}{126}$.