Question 955255: You ask a neighbor to water a sickly plant while you are on vacation. Without water the plant will die with probability 0.85. With water it will die with probability 0.55. You are 82 % certain the neighbor will remember to water the plant.
Find the probability that the plant will die.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! i am not 100% certain, but i think the solution is as follows:
a is the event that the plant will die without water.
b is the event that the plant will die with water.
p(a) = .85
p(b) = .55
c is the event that the neighbor will water the plant.
d is the event that the neighbor will not water the plant.
p(c) = .82
p(d) = 1 - .82 = .18
your expected value is p(a) * p(d) + p(b) * p(c)
this becomes expected value = .85 * .18 + .55 * .82
expected value becomes .604
the justification for this is shown below:
assume the person goes away 1000 times.
82% of the time the neighbor will water the plant and 18% of the time the neighbor won't water the plant.
82% of 1000 = 820 times that the neighbor will water the plant.
18% of 1000 = 180 times that the neighbor will not water the plant.
within the 820 times, the plant will die 55% of the time.
.55 * 820 = 451
451 times out of the 820 times, the plant will die.
within the 180 times, the plant will die 85% of the time.
.85 * 180 = 153
153 times out of the 180 times, the plant will die.
so for the 1000 times, the plant died 451 + 153 = 604 times.
604 / 1000 = .604.
it looks like the solution is good, but i still won't swear to it even though i'm pretty sure it is good.
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