Probability Question: 3-Digit Combination Lock

[MeesaWorldWide]

Homework Statement

A combination lock opens with the correct 3-digit code. Each wheel rotates through the digits 1 to 8. What is the probability of a randomly chosen passcode having exactly 2 fives?

Relevant Equations

n/a

A passcode having 2 fives implies that order does not matter (5 7 5 is the same passcode as it would be if we switched the 2 fives). There are 8 available digits in total, but 5 is being used twice, so we only have 7 options for the third number in the code. There are 3 possible cases for the distribution of the fives ( 5 _ 5 , 5 5 _ , and _ 5 5), so 7 x 3 is 21.
There are 8 x 8 x 8 total possible passcodes (which = 512)
So 21/512 = 0.041 = 4.1%
However, the answer is apparently 8.2% (twice as much), so I'm not sure where I went wrong...?
Any explanation would be greatly appreciated.

 

[FactChecker]

I think I agree with your answer.

A passcode having 2 fives implies that order does not matter (5 7 5 is the same passcode as it would be if we switched the 2 fives). ...

Good. So you have your 3 possibilities for placing the 5's instead of 6. I think a better way to look at it is that there are 3 places to place the non-5 digit.

There are 8 x 8 x 8 total possible passcodes (which = 512)

I agree. I suspect that the book answer use the fact that the order of the 5's doesn't matter here, but I don't think it applies (575 is only counted as one passcode here, not two). This If you divide 512 by 2, you get their answer. I will have to think about it more.

 

[pbuk]

However, the answer is apparently 8.2% (twice as much)

That is not the answer to the question you wrote: your answer is correct.

Can you think of another method to verify your answer?

Spoiler

The probability of choosing five is 1/8

 

[Hornbein]

I also am convinced your answer is correct.