Answer to Question #234577 in Statistics and Probability for Milika

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 breaking the code in one attempt)?
ii) if both codes are implemented, the first followed by the second, what is probability of gaining access in a single attempt?

Answer 1

Number of combinations for the first code: $N_1=10^6=1000000.$

Number of combinations for the second code: $N_2=10^3*24^2=576000.$

i) Since $N_1>N_2$, the first code is better.

ii) $N=N_1N_2=1000000*576000=576000000000=5.76*10^{11}.$

$P=\frac{1}{N}=1.7361*10^{-12}.$