(i) Finding $P(E_1)$ and $P(E_2)$

Step 1 (Definition): Given that 70% of customers pay the first month's bill in time, $E_1$ is the event of paying in time and $E_2$ is the event of not paying in time.

Step 2 (Working):

• $P(E_1) = 70\% = \frac{70}{100} = 0.7$
• $P(E_2) = 1 - P(E_1) = 1 - 0.7 = 0.3$

Step 3 (Final Answer): $P(E_1) = 0.7, P(E_2) = 0.3$ (1 Mark)

(ii) Finding $P(A|E_1)$ and $P(A|E_2)$

Step 1 (Conditional Probability): $A$ is the event of paying the second month's bill in time. Based on the problem statement:

• $P(A|E_1)$ is the probability of paying in time given the first month was paid in time.
• $P(A|E_2)$ is the probability of paying in time given the first month was not paid in time.

Step 2 (Working):

• $P(A|E_1) = 0.8$
• $P(A|E_2) = 0.4$

Step 3 (Final Answer): $P(A|E_1) = 0.8, P(A|E_2) = 0.4$ (1 Mark)

(iii) Finding the probability of customer paying second month's bill in time

Step 1 (Law of Total Probability): By the Law of Total Probability:

Step 2 (Working):

Step 3 (Final Answer): The probability is $0.68$ (2 Marks)

OR (iii) Finding $P(E_1|A)$

Step 1 (Bayes' Theorem):