Question

Pal's gardener is not dependable, the probability that he will forget to water the rose bush is $\frac{2}{3}.$ The rose bush is in questionable condition any how, if watered the probability of its withering is $\frac{1}{2},$ if not watered, the probability of its withering is $\frac{3}{4}.$ Pal went out of station and upon returning, he finds that the rose bush has withered, what is the probability that the gardener did not water the bush.
[Here result is known that the rose bush has withered, therefore. Bayes's theorem should be used]

Answer 1

Let $A=$the event that the rose bush has withered
Let $A_1=$the event that the gardener did not water.
$A_2=$the event that the gardener watered.
By Bayes's theorem required probability,
$P\left(A_1/A\right)=\frac{P\left(A_1\right).P\left(A/A_1\right)}{P\left(A_1\right).P\left(A/A_1\right)+P\left(A_2\right).P\left(A/A_2\right)}$ (i)
Given, $P\left(A_1\right)=\frac{2}{3}$
$\therefore$ $P\left(A_2\right)=\frac{1}{3}$
$P\left(A/A_1\right)=\frac{3}{4},P\left(A/A_2\right)=\frac{1}{2}$
From (i), $P\left(A_1/A\right)=\frac{\frac{2}{3}.\frac{3}{4}}{\frac{2}{3}.\frac{3}{4}+\frac{1}{3}.\frac{1}{2}}=\frac{6}{6+2}=\frac{3}{4}$