Question

You ask your neighbor to water a sickly plant while you are on vacation. Without water, it will die with probability 0.8; with water, it will die with probability 0.15. You are 90 percent 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 the plant is dead upon your return, what is the probability

that your neighbor forgot to water it?

Short answer

P(plant will be alive on return)$=0.785$

P(neighbor forgot to water it given plant is dead on return)$=\frac{16}{43}$

Step-by-step solution

Given information

P(plant dying without water)$=0.8$

P(plant dying with water)$=0.15$

P(neighbor waters the plant)$=0.9$

Assumptions

Let Abe the event plant is alive and Wbe the event that plant is being watered.

Part(a)

$$\begin{gathered} P\left(A\right)=P\left(A\right|W\left)P\right(W)+P(A\left|\overline{W}\right)P\left(\overline{W}\right) \\ \Rightarrow P\left(A\right)=0.85*0.9+0.2*0.1 \\ \Rightarrow P\left(A\right)=0.785 \end{gathered}$$

Part(b)

$$\begin{gathered} P\left(\overline{W}\right|\overline{A})=\frac{P\left(\overline{A}\right|\overline{W}\left)P\right(\overline{W})}{1-P\left(A\right)}=\frac{0.8*0.1}{0.215} \\ \Rightarrow P(\overline{W}\left|\overline{A}\right)=\frac{16}{43} \end{gathered}$$