Q) A departmental store sends bills to charge its customers once a month. Past experience shows that 70% of its customers pay their first month bill in time. The store also found that the customer who pays the bill in time has the probability of 0.8 of paying in time next month and the customer who doesn’t pay in time has the probability of 0.4 of paying in time the next month.

Based on the above information, answer the following questions :
(i) Let E₁ and E₂ respectively denote the event of customer paying or not paying the first month bill in time. Find P(E1), P(E2).
(ii) Let A denotes the event of customer paying second month’s bill in time, then find P(A|E₁) and P(A|E2).
(iii) Find the probability of customer paying second month’s bill in time.
OR
(iii) Find the probability of customer paying first month’s bill in time if it is found that customer has paid the second month’s bill in time.

Ans: From the data given in question, we gather:

1st month paying probability = 0.7

2nd month paying probability = 0.8

3rd month paying probability = 0.4

(i) Calculating P(E1) and P(E2):

Probability of customer paying the first month bill in time P(E1) = 0.7

Probability of customer not paying the first month bill in time P(E2) = 1- 0.7 = 0.3

(ii) Calculating P(A|E₁) and P(A|E2):

It is given that 80% customers pay timely in 2nd month who paid timely in 1st month

∴ P(A|E1) = 0.8

Also it is given that 40% customers pay timely in 2nd month who did not pay timely in 1st month

∴ P(A|E2) = 0.4

(iii) Probability of customer paying second month’s bill in time:

probability of customer paying 2nd month’s bill in time = probability of customer paying 2nd month’s bill in time from the customers who paid 1st bill in time + probability of customer paying 2nd month’s bill in time from the customers who did not pay 1st bill in time

P(A) = P (E1) x P(A|E1) + P (E2) x P(A|E2)

Bu substituting the values from above we get:

P(A) = 0.7 x 0.8 + 0.3 x 0.4 = 0.56 + 0.12

∴ P(A) = 0.68

(iv) Probability of customers timely paying 1st month bill from customers paid the second month’s bill in time

This probability can be calculated by:

or it can be expressed as (using Bayes’ theorem): P(E1|A) = \frac{0.8 \times 0.7}{0.68} = \frac{0.56}{{0.68} = \frac{14}{17}\$

∴ P(E1|A) = 0.8235

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