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this one is a tuffy

rgrocks

New member
Joined
Oct 9, 2009
Messages
18
You ask your neighbor to water a sickly plant while you are on vacation. Without water it will die with a probability .8; with water it will die with probability .15. You are 90% certain that your neighbor will remember to water the plant.
a) what is the probability that the plant will be alive when you return
b) if it is dead what is the probability your neighbor forgot to water the it

this problem seems to have alot of variables that i couldnt firgure out how to put it together
i started creating events like

A-event that the plant will die P(A) is the probability that the plant will die so to find the answer to "a" i have to subtract that value from 1.

but i really dont even know the first step to takling this problem.

if you can help me please describe what you were thinking when you were solving this problem so i can follow along with the steps so i can learn from it. Thanks
 

wjm11

Senior Member
Joined
Nov 13, 2004
Messages
1,417
You ask your neighbor to water a sickly plant while you are on vacation. Without water it will die with a probability .8; with water it will die with probability .15. You are 90% certain that your neighbor will remember to water the plant.
a) what is the probability that the plant will be alive when you return
b) if it is dead what is the probability your neighbor forgot to water the it
These questions can easily be solved if we draw a probability “tree” diagram. Start with two branches, one labeled “tree gets watered” and the other labeled “tree does not get watered.” On the “tree gets watered” branch, write “.9” (the probability it gets watered). On the other branch, write “.1”.

Now draw two more branches off the ends of each branch (4 total). Label them “tree lives: .85”; “tree dies: .15” (off of the “tree gets watered” branch) and “tree lives: .2”; “tree dies: .8” (off of the “tree does not get watered” branch).

Next, multiply the numbers along each path and write the answer off the end of that path. For example, along the “tree gets watered” and “tree lives” path, we have (.9)(.85) = .765. The four probabilities are .765, .135, .02, .08. [Note: the four answers must add up to 1.]

There is a path in which the tree does not get watered but still lives. Add both “tree lives” probabilities together to get the answer to part a): .765 + .02 = .785.

For part b), consider only the paths in which the tree dies. Add the two “tree dies” probabilities together to get the “tree dies” prob: .135 + .08 = .143. So, the probability that the tree did not get watered and died is that particular branch over the total of the condition:

.08/.143 = .559 (approx.)
 

rgrocks

New member
Joined
Oct 9, 2009
Messages
18
wow... why hasnt anyone ever thought me that method.
 
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