Answer to Question #264167 in Statistics and Probability for Faru

Question

a factory produces electric bulbs by three plants: plants-1 ,plants-2, plants-3, produces 3000, 4000 and 3000 units respectively.past experiences shows that 2% products of plant-1 ,3% products of plant -2, 1% products of plant -3 are defectives. a bulb is selected randomly and found defective. find the probability that it is produced by plant -1

Answer 1

Let $A$ denote the event "electric bulb is produced by plant -1".

Let $B$ denote the event "electric bulb is produced by plant -2".

Let $C$ denote the event "electric bulb is produced by plant -3".

Let $D$ denote the event "electric bulb is defective".

$$3000+4000+3000=10000$$

Given

$$P(A)=0.3, P(B)=0.4, P(C)=0.3,$$

$$P(D|A)=0.02, P(D|B)=0.03, P(D|C)=0.01$$

By Bayes' Rule

$$P(A|D)$$

$$=\dfrac{P(A)P(D|A)}{P(A)P(D|A)+P(B)P(D|B)+P(C)P(D|C)}$$

$$=\dfrac{0.3(0.02)}{0.3(0.02)+0.4(0.03)+0.3(0.01)}=\dfrac{2}{7}$$

The probability that electric bulb is produced by plant -1 given that it is defective is $\dfrac{2}{7} .$