Answer to Question #234938 in Statistics and Probability for Nay

Question

(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?

c) A certain Mobile Money Agency is an agent for Zanaco Express, Airtel, MTN and Zamtel Mobile Money. Experience has shown that the probabilities of finding a float for withdrawals of amounts K5000 and above are given as 0.4, 0.5, 0.3 and 0.2, respectively. Experience has also shown that among his clients, the probabilities that a client requires a service from Zanaco Express, Airtel, MTN, Zamtel Mobile Money are given as 0.3, 0.4, 0.2 and 0.1, respec- tively. Assume that each client visits the agency for withdrawals from only one at a time.

What is the probability that a client who wishes to withdraw K5000 will not be successful?

ii) What is the probability that an MTN client who wishes to withdraw K5000 will not be successful?

Answer 1

(i)

If repeating is allowed than you can generate $10^6$ passwords of length $6$ with $10$ digits.

Hence probability of gaining access in a single attempt is

$$p=\dfrac{1}{10^{6}}=10^{-6}$$

If repeating is allowed than you can generate $10^3\cdot(26-2)^2$  passwords of length $3+2.$

Hence probability of gaining access in a single attempt is

$$p=\dfrac{1}{10^3\cdot(26-2)^2}=1.736\times10^{-6}$$

The first code is code is better at preventing unauthorized access.

(ii)

If both codes are implemented, the first followed by the second you can generate $10^6\cdot10^3\cdot(26-2)^2=5.76\cdot10^{11}$ passwords.

Hence probability of gaining access in a single attempt is

$$p=\dfrac{1}{5.76\cdot10^{11}}=1.736\times10^{-12}$$

c)

Consider the following events:

D: withdraw k5000 will not be successful,

ZE: a service from Zanaco Express,

A: a service from Airtel,

MTN: a service from MTN,

ZMM: a service from zamtel mobile money.

i) By the Law of Total Probability

$$P(D)=P(ZE)P(D|ZE)+P(A)P(D|A)$$

$$+P(MTN)P(D|MTN)+P(ZMM)P(D|ZMM)$$

$$=0.3(1-0.4)+0.4(1-0.5)+0.2(1-0.3)$$

$$+0.1(1-0.2)=0.6$$

ii)

$$P(D|MTN)=1-0.3=0.7$$