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AP EAMCET · Maths · Probability

A manufacturing company of bulbs has 3 units \(\mathrm{A}, \mathrm{B}\) and C which produce \(25 \%\), \(35 \%\) and \(40 \%\) of the bulbs respectively. Out of the bulbs produced by A, B, C units, \(5 \%, 4 \%\) and \(2 \%\) are defective respectively. If a bulb is chosen at random and found to be defective, then the probability that it is produced by unit B is

  1. A \(\frac{28}{69}\)
  2. B \(\frac{28}{71}\)
  3. C \(\frac{29}{67}\)
  4. D \(\frac{25}{69}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{28}{69}\)

Step-by-step Solution

Detailed explanation

\( P(D) = P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C) \) \( P(D) = (0.25)(0.05) + (0.35)(0.04) + (0.40)(0.02) \) \( P(D) = 0.0125 + 0.0140 + 0.0080 = 0.0345 \) \( P(B|D) = \frac{P(D|B)P(B)}{P(D)} \) \( P(B|D) = \frac{(0.35)(0.04)}{0.0345} \) \( P(B|D) = \frac{0.0140}{0.0345} \)…