The correct answer is:
A. 24 out of 504.
The probability that the lock code consists of all even digits is 24 out of 504, if the same digit is not used more than once in the lock code.
[tex]|Huntrw6|[/tex]
Option A: 24 out of 504
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 is 24 out of 504.
Given that:
size of lock code is 3
lock code consists of non-zero digits
same digits can't be used again
Since zero can't be used, only 1,2,3,4,5,6,7,8,9 can be used.
By the rule of product in combinatorics, we have, in total, [tex]9 \times 8 \times 7 \: \rm ways = 504 \: \rm combinations[/tex]
The favorable case needs all digits to be even, which means only 2,4,6,8 can be used. The total number of combinations for such lock codes are:
[tex]4 \times 3 \times 2 = 24 \: \rm combinations[/tex]
The probability of getting lock codes with only even digits can be thus calculated as:
[tex]\begin{aligned} P(\text{Only even digits in code}) &= \dfrac{\text{Total favorable cases}}{\text{Total cases' count}}\\&= \dfrac{24}{504}\\\end{aligned}\\[/tex]
Thus, 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 is 24 out of 504
Thus, option A: 24 out of 504 is correct.
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