Answer to Question #234873 in Statistics and Probability for Nay

Question

A consumer agency randomly selected 1700 flights for two major airlines, A and B. The following table gives the two-way classification of these flights based on airline and arrival time. Note that ”less than 30 minutes late” in- cludes flights that arrived early or on time.

b) 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)?

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

Answer 1

Full problem

a) P( more than 1 hour late or airline A)

= ( number of flights of airline A + number of flights of airline B which are more than 1 hour late ) / 1700

$= \frac{429 + 390 + 92 + 80}{1700} = 0.5829$

Therefore 0.5829 is the required probability here.

b) P ( airline B or less than 30 min late )

= ( number of flights of airline B + number of flights with airline A that are less than 30 min late ) / 1700

$= \frac{393 + 316 + 80 + 429}{ 1700} = 0.7165$

Therefore 0.7165 is the required probability here.

c) P( airline A or Airline B) = 1

since only 2 airlines are there.

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?

Solution:

(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!

(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 $10^9$ multiplied by 242. So $5.76 \times 10^{11}$ combinations. In one attempt a person can only try one combination, so the probability that this combination is the correct code is $\frac{1}{(5.76 \times10^{11})}$ which is equal to $1.736\times 10^{-12}$ . 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.