Question

Ram'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 anyhow if watered the probability of its withering is $\frac{1}{2}$, if not watered, the probability of its withering is $\frac{3}{4}$. Ram went out of the station and upon returning, he finds that the rosebush has withered. The probability that the gardener did not water the bush is?

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{3}{4}$
• D. none of the above

Answer 1

The correct option is C

$\frac{3}{4}$

Explanation for the correct option:

Let A be the event where ram's gardener forgot to water the rose bush

Let B be the event where ram's gardener watered the rose bush

Let C be the event where the rose bush withered

Given:

$P\left(A\right)=\frac{2}{3}$

$P\left(C\right|B)=\frac{1}{2}$

$P\left(C\right|A)=\frac{3}{4}$ So,

$P\left(B\right)=1-P\left(A\right)$

$$\begin{gathered} =1-\frac{2}{3} \\ =\frac{1}{3} \end{gathered}$$

To find: $P\left(A\right|C)$

By Bayes theorem:

P(A|C)$=\frac{P\left(C\right|A).P(A)}{P\left(C\right)}$ …….(i)

we need to find $P\left(C\right)$

by using the law of total probability which states that if $B_n$ are pairwise disjoint events then for any event A of the same probability space:

$\underset{n}{\overset{}{P\left(A\right)=\sum }}P\left(A\right|B_n).P(B_n)$

therefore

$P\left(C\right)=P\left(C\right|A).P(A)+P(C\left|B\right).P\left(B\right)$

$=\left(\frac{3}{4}\right)\left(\frac{2}{3}\right)+\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)$

$=\frac{2}{3}$

we know from (i) that:

$P\left(A\right|C)=\frac{P\left(C\right|A).P(A)}{P\left(C\right)}$

substituting the values

$P\left(A\right|C)=\frac{\left({\displaystyle \frac{3}{4}}\right)\left({\displaystyle \frac{2}{3}}\right)}{{\displaystyle \frac{2}{3}}}$

$=\frac{3}{4}$

Hence, option (C) is the correct answer