SOLUTION: 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

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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