Answer to Question #231651 in Statistics and Probability for Jamilah

Question

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, respectively. 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?
[2 marks

Answer 1

i) Using law of total probability, we simply find the sum product of the probabilities of being a client with their success probabilities to get the required total success probability here as:

$= 0.4*0.3 + 0.5*0.4 + 0.3*0.2 + 0.2*0.1\\ = 0.12 + 0.2 + 0.06 + 0.02 \\ = 0.4$

Therefore 0.4 is the required probability here.

ii) P( not successful | MTN) = 1 - 0.3 = 0.7 is the required probability here.

iii) P( not successful | Airtel ) = 1 - 0.5 = 0.5 is the required probability here.

iv) Given that a client is not successful, probability that he / she is from Zamtel is computed here using Bayes theorem as:

= $\frac{0.2*(1 - 0.1) }{ (1 - 0.4)}$

Note that the denominator here is 1- 0.4 as 0.4 is the probability of being successful

= $\frac{0.18 }{ 0.6}$

= 0.3 is the required probability here.