Question

On every evening, a student either watches TV or reads a book. The probability of watching TV is \( \frac{4}{5} \). If he watches TV, the probability that he will fall asleep is \( \frac{3}{4} \), and it is \( \frac{1}{4} \) when he reads a book. If the student is found to be asleep on an evening, the probability that he watched the TV is:

• A. \( \frac{11}{13} \)
• B. \( \frac{12}{13} \)
• C. \( \frac{2}{13} \)
• D. \( \frac{4}{13} \)

Hint:

To find conditional probability when outcomes can arise from multiple events, use Bayes’ Theorem with total probability in the denominator.

Answer 1

The Correct Option is B

Let: \[ \begin{aligned} P(T) &= \frac{4}{5},
P(R) = \frac{1}{5}
\text{(complement)}
P(A|T) &= \frac{3}{4},
P(A|R) = \frac{1}{4} \end{aligned} \] Using total probability theorem: \[ P(A) = P(T) . P(A|T) + P(R) . P(A|R) = \frac{4}{5} . \frac{3}{4} + \frac{1}{5} . \frac{1}{4} = \frac{12}{20} + \frac{1}{20} = \frac{13}{20} \] Now apply Bayes' Theorem to find \( P(T|A) \): \[ P(T|A) = \frac{P(T) . P(A|T)}{P(A)} = \frac{\frac{4}{5} . \frac{3}{4}}{\frac{13}{20}} = \frac{12}{20} \div \frac{13}{20} = \frac{12}{13} \]