Answer to Question #153699 in Statistics and Probability for Shiv

Question

A phone lock is a combination of four digits where each digit can vary from 0 to 9. If a person tries any combination randomly what is the probability that he puts at least two digits right but is unable to unlock the phone?

Answer 1

The total number of combinations is 10⁴ = 10000.

The number of combinations in which each digit is incorrect, equals to 9⁴=6561.

The number of combinations in which only the first (only the second, etc) digit is correct, equals to 9³=729, therefore, the number of combinations in which only one digit is correct, equals to 729*4 = 2916.

The number of combinations in which less than 2 digits are incorrect, equals to 6561+2916 = 9477.

The number of combinations in which at least 2 digits are correct, equals to 10000 - 9477 = 523, and 522 of them are wrong.

The probability that a person puts a wrong combination with at least 2 correct digits, is equal to 522/10000 = 0.0522 = 5.22%

Answer. 5.22%