Answer to Question #234780 in Statistics and Probability for Booblex

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)?

[3 marks]

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

Answer 1

Assumptions

We are not told whether digits/letters can be repeated or not. So we will assume repetition is allowed. Similarly, it is possible for the first digit to be zero since a code is needed and not a number. Thus, a code can be one or more zeros.

i. Better code

The fist code has six digits chosen from 10 numbers (0-9). Thus, there are 10 ways of choosing each of the six digits.

The total number of possible codes is $10^6=1,000,000$

The probability of breaking the code in one attempt is $\frac{1}{10^6}=\frac{1}{1,000,000}$

The second code has three digits selected from 0-9 and two letters selected from A-Z excluding and O. Assuming repetition is allowed, there are 10 ways of selecting each digit and 24 ways of selecting each letter.

The total number of possible codes is $10^3\times24^2=576,000$

The probability of breaking the code in one attempt is $\frac{1}{576,000}$

Since $\frac{1}{1,000,000}<\frac{1}{576,000}$ the first code is better in preventing unauthorized access.

ii

If both codes are implemented, (independent codes) the probability of gaining an authorized access in a single attempt is $\frac{1}{1,000,000}\times\frac{1}{576,000}=\frac{1}{576,000,000,000}$