Answer to Question #234484 in Statistics and Probability for chisenga

Question

Two new computer codes are being developed to prevent unauthorized
access to classified information. The first consists of six digits (each chosen
from 0 to 9); the second consists of three digits (from 0 to 9) followed by two
letters (A to Z, excluding I and O).

(ii) If both codes are implemented, the first followed by the second, what is
probability of gaining access in a single attempt?

Answer 1

(i) Code1: It has 6 digits chosen from 0 to 9. So the total available combinations are 10⁶. In one attempt a person can only try one combination, so the probability that this combination is the correct code is 1/10⁶ = 10^-6

Code 2: The second code has 3 digits taken from 0 to 9, and 2 letters taken from 24 alphabets(Since I and O are excluded) . So total available combinations are 10³ multiplied by 24². So 576,000 combinations. In one attempt a person can only try one combination, so the probability that this combination is the correct code is 1/576000 which is equal to 1.736*10^-6

Since the probability for breaking the code in one try is more for code2, Code1 is the better code for preventing unauthorized access!

(ii) If both codes are implemented the first followed by the second, there will be 9 digits and 2 letters in the new code. So the total available combinations are 10⁹ multiplied by 24². So 5.76*10¹¹ combinations. In one attempt a person can only try one combination, so the probability that this combination is the correct code is 1/(5.76*10¹¹) which is equal to 1.736*10^-12. As expected, this is really small when compared to the other two probabilities that we found out because this code is longer and is harder to break.