Question

Initially you have two working phones: an older "2+1" smartphone and a newer "mexus 6". Unfortunately, during a peaceful dinner with your family, the beautiful "mexus 6" hits the floor, so its screen is completely shuttered, and the phone is unusable. For each phone you have several options in mind. For "2+1", you can either do nothing (keep it) or sell it for $100. Since your "mexus 6" was purchased within last 30 days, you are still eligible for purchasing an insurance that covers drops and leakages, with premium $120 for a year plus $40 additional cost for each accident (insurance deductible is $40 and loss is not completely covered by the insurance payment). Overall, there are several options with "mexus 6": do nothing (throw it away); purchase the insurance and use it to repair the phone; repair without insurance, which costs $150; sell a repaired phone for $250 (when no insurance was purchased); sell a repaired phone with insurance for $300. Also, you may either buy or not a new different phone for $400.
1. Suppose that you are willing to end up with exactly one working phone. How many action scenarios can lead you to that outcome? (A scenario is a combination of decisions regarding three phones: "2+1", "mexus 6", and the new one.)

2.Given that you end up with exactly two working phones (one major, one backup), what is the smallest total cost of the scenario achieving that (i.e., what is the smallest amount of money that you need to spend if you want to have two phones)?

Answer 1

Answer:

1) there are four possible scenarios

2) It will cost 150 to repair the mexus this will be the cheapest options

Explanation:

1) we have to disclosure probabilities

1 keep 2+1 trow mexus

2sale 2+1 repair mexus

3 sale 2+1 trow mexus and purchase a new cellphone

4 sale 2+1 prepair and sale mexus and purchase a new cellphone

2)

repair the mexus with no insurance will be the least expensive option as will only cost 150 dollars.