# Question: A department store that bills its charge account customers once

A department store that bills its charge- account customers once a month has found that if a customer pays promptly one month, the probability is 0.90 that he or she will also pay promptly the next month; however, if a customer does not pay promptly one month, the probability that he or she will pay promptly the next month is only 0.40. (Assume that the probability of paying or not paying on Any given month depends only on the outcome of the previous month.)

(a) What is the probability that a customer who pays promptly one month will also pay promptly the next three months?

(b) What is the probability that a customer who does not pay promptly one month will also not pay promptly the next two months And then make a prompt payment the month after that?

(a) What is the probability that a customer who pays promptly one month will also pay promptly the next three months?

(b) What is the probability that a customer who does not pay promptly one month will also not pay promptly the next two months And then make a prompt payment the month after that?

## Relevant Questions

With reference to Figure 2.15, verify that events A, B, C, And D are independent. The region rep-resenting A consists of two circles, And so do the regions representing B And C. Figure 2.15 With reference to Example 2.25, suppose that we discover later that the job was completed on time. What is the probability that there had been a strike? In exercise The completion of a construction job may be delayed ...A series system consists of two components having the reliabilities 0.98 And 0.99, respectively, connected to a parallel subsystem containing five components having the reliabilities 0.75, 0.60, 0.65, 0.70, And 0.60, ...Given three events A, B, And C such that P(A ∩ B ∩ C) Z0 And P(C| A ∩ B) = P(C| B) , show that P(A| B ∩ C) = P(A| B) . Prove Theorem 2.12 by making use of the following generalization of the distributive law given in part (b) of Exercise 2.1: A ∩(B1. B2. · · · ∪ Bk) =(A ∩ B1) .(A ∩ B2) ∪ · · · ∪ (A ∩ Bk)Post your question