For a delivery service problem, there are four basic outcomes B1, B2, B3 and B4

Question

For a delivery service problem, there are four basic outcomes B1, B2, B3 and B4 where B1 is delivery on day 1 (probability = 0.6). B2 is delivery on day 2(probability = 0.2), B3 is delivery on day 3 (probability = 0.1), and B4 is delivery is after day 3 (probability = 0.1). There is also a possibility that on any day, the delivery could be correct (Event A1) or incorrect (Event A2). For a correct order (event A1), probability of delivery on Day 1 is 0.57, probability of delivery on Day 2 is 0.18, probability of delivery on Day 3 is 0.08, probability of delivery after Day 3 is 0.0.03. Find the following: P (A1) P(A2)middot P(A1/B3) P(A2/B3)

Explanation / Answer

Part 1

The delivery would be correct when delivery is correct on each day. So the probability of delivery being correct on day 1 = 0.57*0.6. Similarly we will find the probability of delivery being correct on each day and then add them up to get the probability of event A1.Therefore we have,

P(A1) = 0.57*0.6+0.0.18*0.2+0.08*0.1+0.03*0.1 = 0.353

Part 2

The probability of delivery being incorrect is the complementary probability of event A1. Therefore we have,

P(A2) = 1 - P(A1 ) = 1 - 0.353 = 0.647

Part 3.

For this part we are required to compute the probability of delivery being correct if it is given that it is day 3.

P(A1|B3) = 0.08

Part 4

For this part we are required to compute the probability that dellivery is incorrect given that it is on day 3.

P(A2|B3) = 1 - 0.08 = 0.92

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.