Question

A dealer bought a fixed number of laptops from 3 different companies A, B and C. Among these laptops, \(28\%\) are bought from A, \(32\%\) are bought from B and \(40\%\) are bought from C. \(2.5\%\) of laptops bought from A, \(1.5\%\) bought from B and \(1\%\) bought from C are likely to be defective. If a customer found that the laptop bought by him is defective, then the probability that it was from B is

• A. \(\dfrac{12}{39}\)
• B. \(\dfrac{49}{158}\)
• C. \(\dfrac{17}{78}\)
• D. \(\dfrac{24}{79}\)

Hint:

For questions involving conditional probability after observing an event, use Bayes' theorem: \[ \boxed{ P(A_i|B) = \frac{P(A_i)P(B|A_i)} {\sum P(A_j)P(B|A_j)} } \]

Answer 1

The Correct Option is D

Step 1: Write the given probabilities. Let \(D\) denote the event that a laptop is defective. \[ P(A)=0.28,\qquad P(B)=0.32,\qquad P(C)=0.40. \] Also, \[ P(D|A)=0.025,\qquad P(D|B)=0.015,\qquad P(D|C)=0.01. \]

Step 2: Find the total probability of a defective laptop. Using the law of total probability, \[ P(D) = P(A)P(D|A) + P(B)P(D|B) + P(C)P(D|C). \] Hence, \[ P(D) = 0.28(0.025) + 0.32(0.015) + 0.40(0.01) = 0.007+0.0048+0.004 = 0.0158. \]

Step 3: Apply Bayes' theorem. \[ P(B|D) = \frac{P(B)P(D|B)}{P(D)} = \frac{0.32\times0.015}{0.0158} = \frac{0.0048}{0.0158}. \] Multiplying numerator and denominator by \(10000\), \[ P(B|D) = \frac{48}{158} = \boxed{\frac{24}{79}}. \] Hence, the correct option is \(\boxed{(D)}\).