Conditional Probability Question: What is the probability that the plant will be alive when you return? Before leaving for a two week trip to Hawaii, you ask a neighbor to water a plant that you have at home. You are 85% certain that the neighbor will remember to water the plant. You also estimate that without water, the plant will die with probability 90%, and with water, it will survive with probability 90%. 
What is the probability that the plant will be alive when you return?

I am thinking I have to solve P[alive]=P[alive|water]*P[water]+P[alive|no water]*P[no water] but I'm not sure if this is the correct interpretation.

I have P[alive|water]=.9; P[water]=.85; P[no water]=.15 but I am having trouble solving P[alive|no water]. I would assume I would need to solve P[alive and no water]/P[no water] but I'm not sure. Any help would be appreciated. Thanks
What you did is correct. Now
$$P(\mbox{alive} | \mbox{no water}) = 1- P(\mbox{die} | \mbox{no water}) = 1 - 0.9 = 0.1$$
and you can conclude the rest.
To produce the probability of plant being alive, use the numbers you have. 

(Certainty of Being Watered)(Plant will survive with Water)+(Certainty of Not being Watered)(Plant surviving without Water)