If repetition is not allowed in the 5 digit pin, then number of possible pins = 10C5 = 252
P(guessing it right in the first attempt) = 1/252
P(guessing it right in the second attempt) = 251/252 1/251 = 1/252
P(guessing it right in the second attempt) = 251/252 250/251 1/250 = 1/252
P(of guessing it right in 3 attempts .with different pin each time) = 1/252 + 1/252 + 1/252 = 3/252 = 1/84 = 0.0119
If repetition is allowed in the 5 digit pin, then number of possible pins = 10^5 = 100000
P(guessing it right in the first attempt) = 1/100000
P(guessing it right in the second attempt) = 99999/100000 1/99999 = 1/100000
P(guessing it right in the second attempt) = 99999/100000 99998/99999 1/99998 = 1/100000
P(of guessing it right in 3 attempts .with different pin each time) = 1/100000 + 1/100000 + 1/100000 = 3/100000 = 0.00003
