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.)