Question

Every morning, MR.$X$ either reads a book or watches T.V. The probability that he watches T.V. is $4/5$. If he watched T.V., there is a probability of $3/4$ that he falls asleep. If he reads a book the probability that he falls asleep is $1/4$. On one evening Mr.$X$ is found to be asleep. The probability that he watched T.V. is:

• A. $\frac{1}{13}$
• B. $\frac{12}{13}$
• C. $\frac{13}{20}$
• D. $\frac{2}{3}$

Answer 1

The correct option is B $\frac{12}{13}$

Let $E_1$ be the event that MR.$X$ reads a book and $E_2$ be the event that he watches T.V.
Let $A$ be the event that Mr.$X$ falls asleep
Given:
$P\left(E_1\right)=\frac{1}{5},P\left(E_2\right)=\frac{4}{5};P\left(A/E_1\right)=\frac{1}{4},P\left(A/E_2\right)=\frac{3}{4}$
$P\left(E_2/A\right)=\left[\frac{P\left(E_2\right).P\left(A/E_2\right)}{P\left(E_1\right).P\left(A/E_1\right)+P\left(E_2\right).P\left(A/E_2\right)}\right]$
$=\frac{(\frac{4}{5}\times \frac{3}{4})}{(\frac{1}{5}\times \frac{1}{4})+(\frac{4}{5}\times \frac{3}{4})}=\frac{12}{13}$