A shopkeeper buys a particular type of electric bulbs from three manufacturers $ M_1, M_2, M_3 $. He buys 25% of his requirement from $ M_1 $, 45% from $ M_2 $, and 30% from $ M_3 $. Based on past experience, he found that 2% of type $ M_1 $ bulbs are defective, whereas only 1% of type $ M_1 $ and type $ M_2 $ are defective. If a bulb chosen by him at random is defective, then the probability that it was of type $ M_3 $ is:
Show Hint
Use Bayes' Theorem to solve probability problems involving conditional probabilities. Be sure to calculate the total probability of the event first.
Let \( A_1, A_2, A_3 \) represent the events that a bulb is chosen from \( M_1, M_2, M_3 \) respectively, and \( D \) be the event that a bulb is defective.
We know the following probabilities:
\( P(A_1) = 0.25 \), \( P(A_2) = 0.45 \), \( P(A_3) = 0.30 \)
\( P(D|A_1) = 0.02 \), \( P(D|A_2) = 0.01 \), \( P(D|A_3) = 0.01 \) (since only 1% of type \( M_3 \) bulbs are defective).
Using Bayes' Theorem:
\[
P(A_3|D) = \frac{P(D|A_3)P(A_3)}{P(D)}
\]
Where \( P(D) \) is the total probability of a defective bulb:
\[
P(D) = P(D|A_1)P(A_1) + P(D|A_2)P(A_2) + P(D|A_3)P(A_3)
\]
Substitute the values:
\[
P(D) = (0.02)(0.25) + (0.01)(0.45) + (0.01)(0.30) = 0.005 + 0.0045 + 0.003 = 0.0125
\]
Now, calculate \( P(A_3|D) \):
\[
P(A_3|D) = \frac{(0.01)(0.30)}{0.0125} = \frac{0.003}{0.0125} = \frac{6}{13}
\]
Thus, the probability that the defective bulb is of type \( M_3 \) is \( \boxed{\frac{6}{13}} \).
