Question

You ask your neighbor to water a sickly plant while you are on vacation. Without water, it will die with probability .$8$; with water, it will die with probability .$15$. 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

a). The probability that the plant will be alive when you return is $0.785$.

b). The probability that your neighbor forgot to water it islocalid="1650015942117" $0.3721$.

Step-by-step solution

 Given Information (Part a) 

$W$-. the plants were watered

$P\left(D\mid W^c\right)=0.8$

$P(D\mid W)=0.15$

$P\left(W\right)=0.9$

Explanation (Part a)

Formula of total probability can be applied here (because $W\cap W^c=\varnothing ,P\left(W\cup W^c\right)=1$ ):

$P\left(D\right)=P\left(D\mid W^c\right)P\left(W^c\right)+P(D\mid W)P\left(W\right)$

Formula for complement $\Rightarrow P\left(W^c\right)=1-P\left(W\right)=0.1$

Substitution of familiar probabilities:

$P\left(D\right)=0.8·0.1+0.15·0.9=0.215$

Now formula for complement applied again:

$P\left(D^c\right)=1-P\left(D\right)=1-0.215=0.785$

Given Information (Part b)

$W$-. the plants were watered

$D$- the plants died

Given probabilities:

$P\left(D\mid W^c\right)=0.8$$P(D\mid W)=0.15$$P\left(W\right)=0.9$

Explanation (Part b)

The definition of conditional probability gives:

$P\left(W^c\mid D\right)=\frac{P\left(W^c\cap D\right)}{P\left(D\right)}\text{and}P\left(D\mid W^c\right)·P\left(W^c\right)=P\left(W^c\cap D\right)$

Which can be fused into:

$P\left(W^c\mid D\right)=\frac{P\left(D\mid W^c\right)·P\left(W^c\right)}{P\left(D\right)}$

In part a) we computed $P\left(W^c\right)=0.1,P\left(D\right)=0.215$is stated in the beginning, therefore:

localid="1650015923077" $P\left(W^c\mid D\right)=\frac{0.8·0.1}{0.215}=\frac{16}{43}\approx 0.3721$