Question

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:

1. 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).
2. Let A denotes the event of customer paying second month’s bill in time, then find P(A|E₁) and P(A|E₂).
3. Find the probability of customer paying second month's bill in time.
                                            OR
   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. 

Answer 1

Given, 70% customers pay their first month bill in time, therefore, the customers who do not pay bill in time is 30%.

(i) E₁ and E₂ represent customer paying or not paying the first month bill on schedule.

Then, P(E₁) = 70% = 0.7

and P(E₂) = 30 % = 0.3

(ii) Given that there is a 0.8 chance that customers will pay their bills on time the following month and a 0.4 chance that they won't. 

Let A represent the customer's timely payment of the second month's bill.

∴ `P(A/E_1) = P` (customer paying second month bill in time when they pay first month bill)

= 0.8

and `P(A/E_2) = P` (customer paying second month bill in time when they do not pay first month bill in time)

= 0.4

(iii) The probability that the customer will pay the bill for the second month on time is equal to 0.7 × 0.8 = 0.56.

                                   OR

If it turns out that the customer paid the second month's bill on time, the probability that they paid the first month's bill on time is `P(E_1/A)`.

`P(E_1/A) = (P(E_1) · P(A/E_1))/(P(E_1) · P(A/E_1) + P(E_2) · P(A/E_2))`

= `(0.7 xx 0.8)/(0.7 xx 0.8 + 0.3 xx 0.4)`

= `0.56/(0.56 + 0.12)`

= `0.56/0.68`

= 0.824