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Sir there is a question that
Scrap value of machine 2 is 120000.
It is felt that final scrap value of machine depends on two factors: whether or not a new supplier enters the market (which would reduce likely scrap value) & strength of dollar against other currencies (since sales of used machines will be made abroad & invoiced in foreign currency). Adverse effects will each reduce scrap value by 10% of fig used in investment appraisal. Relevant probabilities areNew supplier Prob Strong $ Prob
Yes 0.4 yes 0.3
No 0.6 no 0.7What is EV of scrap proceeds from machine 2
In solution they have done as follows
Scrap value Prob EV
120000 0.6×0.7 = 0.42 50400
108000 0.4×0.7+0.6×0.3=0.46 49680
96000 0.4×0.3=0.12 11520Can you please explain me the above solution
If neither of the adverse effects happen, then the scrap will be 120,000.
If just one of the adverse effects happen then the scrap will be 90% x 120,000 = 108,000
If both of the adverse effects happen, then the scrap will be 120,000 x 80% = 96,000
The probability of no new supplier is 0.6, and the probability of there being no problem with the exchange rate if 0.7.
Therefore the expected value of no adverse effects is 120,000 x 0.6 x 0.7.
For there to be just one adverse effect (and the other effect to be OK):
For adverse new supplier but not adverse exchange rate is 0.4 x 0.7
For not adverse new supplier but adverse exchange rate is 0.6 x 0.3Therefore the probability of just one adverse effect is (0.4 x 0.7) + (0.6 x 0.3)
For two adverse effects, then probability is 0.4 x 0.3
For Kaplan to include this example is a bit silly of them. OK, it is basic probabilities, but FM is not a probability exam. The chances of something like this being asked in the exam are virtually zero, and even if it was asked there is no way that it could be as much as 50% of a question C and there would be enough elsewhere in the question to still be able to pass it.
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