(B) Let $E$ be the event that the chosen bulb is defective. Let $M_1, M_2, M_3$ be the events that the bulb is purchased from manufacturers $M_1, M_2,

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(B) Let $E$ be the event that the chosen bulb is defective. Let $M_1, M_2, M_3$ be the events that the bulb is purchased from manufacturers $M_1, M_2, M_3$ respectively.Given probabilities:$P(M_1) = 0.25, P(M_2) = 0.45, P(M_3) = 0.30$$P(E|M_1) = 0.01, P(E|M_2) = 0.01, P(E|M_3) = 0.02$Using Bayes' Theorem,the probability that the defective bulb is from $M_3$ is:$P(M_3|E) = \frac{P(M_3) \cdot P(E|M_3)}{P(M_1) \cdot P(E|M_1) + P(M_2) \cdot P(E|M_2) + P(M_3) \cdot P(E|M_3)}$$P(M_3|E) = \frac{0.30 \times 0.02}{(0.25 \times 0.01) + (0.45 \times 0.01) + (0.30 \times 0.02)}$$P(M_3|E) = \frac{0.006}{0.0025 + 0.0045 + 0.0060} = \frac{0.006}{0.0130} = \frac{6}{13}$

Class 12 Mathematics Probability Baye's theorem

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$A$ shopkeeper buys a particular type of electric bulbs from three manufacturers $M_1, M_2$,and $M_3$. 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 $M_3$ bulbs are defective,whereas only $1 \%$ of type $M_1$ and type $M_2$ bulbs are defective. If a bulb chosen by him at random is defective,then the probability that it was of type $M_3$ is

AP EAMCET 2023 All AP EAMCET

A
$\frac{5}{13}$
B
$\frac{6}{13}$
C
$\frac{7}{13}$
D
$\frac{8}{13}$

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