Easyelectrics supermarket orders light bulbs from suppliers, aa electronics and aaa electronics. easyelectrics purchases 30% of its light bulbs from aa and 70% from aaa. two percent of the light bulbs purchased from aa are defective, while three percent of the bulbs from aaa are defective. find the probability that a randomly selected light bulb a) was purchased from aa and is defectivie b) is defective

(a) 0.006

               (b) 0.027

Step-by-step explanation:

Given : P(AA) = 0.3 and P(AAA) = 0.70

Let event that a bulb is defective be denoted by D and not defective be D';

Conditional probabilities given are :

P(D/AA) = 0.02 and P(D/AAA) = 0.03

Thus P(D'/AA) = 1 - 0.02 = 0.98

and P(D'/AAA) = 1 - 0.03 = 0.97

(a) P(bulb from AA and defective) = P ( AA and D)

                                                       = P(AA) x P(D/AA)

                                                       = 0.3 x 0.02 = 0.006

(b) P(Defective) = P(from AA and defective) + P( from AAA and defective)

                         = P(AA) x P(D/AA) + P(AAA) x P(D/AAA)

                         = 0.3(0.02) + 0.70(0.03)

                         = 0.027

1.3 million hoops i believe