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 elther.buy or not a new different phone for $400.

Answer 1

For 2+1 you have two options either to keep it or sell it. To be with exactly one phone when you decide to keep 2+1 you have to sell mexus or not get it repaired(just throw it) ,selling also have two option either selling with insurance or without after repairs hence if you keep 2+1 you have 3 options for mexus(sell with or without insurence or throw it)

If you decide to sell 2+1 you have to keep mexus or buy new phone so as to have one workable phone keeping mexus has two option repair with or without insurance, buying anew phone gives me 3 options to do away with mexus( throw it, sell with or without insurance) so if we sell 2+1 we have 2+3= 5 option

Total option =5(selling 2+1) +3(keeping 2+1)=8

For two phones smallest cost will come when we keep 2+1 since it is already worked for another alternative lowest cost will be when you repair mexus without insurance hence incurring cost of 150 .