You go on vacation and ask your friends to water your plant. If your friends water the plant, it will survive your absence with probability $0.9$. Otherwise, if they forget to water it, it will die with probability $0.6$. Your friends will forget to water your plant with probability $0.3$. You return home from your vacation and find your plant dead. What is the probability that your friends did not water it?
Solution attempt: Let A denote event your friends water the plant.
Let B denote event the plant dies.
We have $P(B^C|A)=0.9, P(B|A^c)=0.6,P(A^c)=0.3$
We want to calculate $P(A^c|B) = \frac{P(B|A^c) \cdot P(A^c)}{P(B|A^c) \cdot P(A^c)+P(B|A)P(A)} = \frac{0.6 \cdot 0.3}{0.6 \cdot 0.3 + (1-0.9) \cdot (1-0.3)} = 0.72$.
Is this correct?
Here +D is Dead , -D is not Dead
+W is Watered , -W is not Watered
Listing the various Probabilities given :
$P(+D|+W)=0.1$
$P(+D|-W)=0.6$
$P(-D|+W)=0.9$
$P(-D|-W)=0.4$
$P(+W)=0.7$
$P(-W)=0.3$
Evaluating the Probabilities necessary :
$P(+D)=P(+D|+W)P(+W)+P(+D|-W)P(-W)$ (Summation over Dead with Watering + Dead without Watering)
$P(+D)=0.1 \times 0.7 + 0.6 \times 0.3$
$P(+D)=0.07 + 0.18$
$P(+D)=0.25$
$P(-D)=0.75$
Computing the wanted Probability :
$P(-W|+D)=P(+D|-W)P(-W)/P(+D)$
$P(-W|+D)=0.6 \times 0.3 / 0.25$
$P(-W|+D)=18/25$
$P(-W|+D)=0.72$
Your Solution has been verified !