Answer to Question #234482 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).

(i) Which code is better at preventing unauthorized access (defined as break
ing the code in one attempt)?

Answer 1

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

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 $10^3$ 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 $\frac{1}{576000}$ which is equal to $1.736 \times 10^{-6}$

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