Suppose we are given three boxes, Box A contains 20 light bulbs, of which 10 are defective, Box B contains 15 light bulbs, of which 7 are defective and Box C contains 10 light bulbs, of which 5 are defective. We select a box at random and then draw a light bulb from that box at random. (a) What is the probability that the bulb is defective? (b) What is the probability that the bulb is good?​

0.49

0.51

Explanation:

Probability that bulb is defective :

Let :

b1 = box 1 ; b2 = box 2 ; b3 = box 3

d = defective

P(defective bulb) = (p(b1) * (d|b1)) + (p(b2) * p(d|b2)) + (p(b3) * p(d|b3))

P(defective bulb) = (1/3 * 10/20) + (1/3 * 7/15) + (1/3 * 5/10))

P(defective bulb) = 10/60 + 7/45 + 5/30

P(defective bulb) = 1/6 + 7/45 + 1/6 = 0.4888

= 0.49

P(bulb is good) = 1 - P(defective bulb) = 1 - 0.49 = 0.51

answer:

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