Answer to Question #234549 in Statistics and Probability for Light

Question

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.

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

[3 marks]

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

[2 marks]

(iii) What is the probability that an Airtel client who wishes to withdraw K5000 will not be successful?

[2 marks]

(iv) A certain client has just failed to make a withdrawal of K7000, what is the probability that he/she is a Zamtel Mobile money client?

Answer 1

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)

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

iii)

$$P(D|A)=1-0.5=0.5$$

iv) By Bayes' Theorem

$$P(ZMM|D)=\dfrac{P(D|ZMM)P(ZMM)}{P(D)}$$

$$=\dfrac{(1-0.2)(0.1)}{0.6}=\dfrac{2}{15}\approx0.1333$$