Question 468417
there are 6 digits from 2 through 7.
they are:   {2,3,4,5,6,7}
The number of possible access codes are 6^3 = 216
The probability of randomly selecting the correct access code is (1/6)^3 = 1/216.
The probability of not selecting the correct code is 1 - 1/216 = 215/216.
I can't show you all the codes because there are too many.
I can, however, show you a much simpler situation.
suppose the possible digits are 3 and 4.
That's 2 possible selections per digit.
With repetition of digits, the number of possible combinations are 2^3 = 8
Those possible combinations are:
111
112
121
122
211
212
221
222
The probability of randomly selecting the correct code is (1/2)^2 which becomes 1/8.
Assume the code is 211.
You can see that there is only 1 code out of the 8 possible codes that contains 211.
The probability of selecting 211 randomly is therefore 1 out of 8.
The probability of NOT selecting the correct code is 1 - 1/8 = 7/8.
You can see that there are 7 out of 8 codes that do not contain 211.
With 6 possible numbers per digit, the number of possibilities is much greater and the probability of guessing the correct code is also much smaller.