Question

A computer lab has two printers handling certain types of printing jobs. Printer-I and Printer-II handle 40% and 60% of the jobs, respectively. For a typical printing job, printing time (in minutes) of Printer-I follows $N(10, 4)$ and that of Printer-II follows $U(1, 21)$. If a randomly selected printing job is found to have been completed in less than 10 minutes, then the conditional probability that it was handled by the Printer-II equals .......... (round off to two decimal places).

Hint:

Bayes' theorem allows conditional probabilities to be computed by comparing likelihoods weighted by prior probabilities.

Answer 1

Correct Answer: 0.55 - 0.6

Step 1: Use Bayes' theorem. 
\[ P(\text{II} | T < 10) = \frac{P(T < 10 | \text{II}) P(\text{II})}{P(T < 10 | \text{I}) P(\text{I}) + P(T < 10 | \text{II}) P(\text{II})}. \]

Step 2: Compute each term. 
For Printer-I: $T \sim N(10,4)$ ⇒ $P(T < 10) = P(Z < 0) = 0.5.$ For Printer-II: $T \sim U(1,21)$ ⇒ $P(T < 10) = \frac{10 - 1}{21 - 1} = \frac{9}{20} = 0.45.$ \[ P(\text{II} | T < 10) = \frac{0.45 \times 0.6}{0.5 \times 0.4 + 0.45 \times 0.6} = \frac{0.27}{0.48} = 0.5625. \] 

Answer: $\boxed{0.56}$