A chain of electronics stores sells three different brands o

A chain of electronics stores sells three different brands of Blu-ray players. Fifty percent of its sales are brand 1 players (the least expensive), 30% are brand 2, and 20% are brand 3 (the most expensive). Each manufacturer offers a one-year warranty on parts and labor. It is known that 25% of brand 1’s players require warranty repair work, whereas the corresponding percentages for brands 2 and 3 are 20% and 10%, respectively.

(i) What is the probability that a randomly selected purchaser has bought a brand 1 player that will need repair while under warranty?

(ii) What is the probability that a randomly selected purchaser has a player that will need repair while under warranty?

(iii) If a customer returns to the store with a player that needs warranty repair work, what is the probability that it is a brand 1 player? A brand 2 player? A brand 3 player?

Solution

Let

B1, B2, B3 = the respective brands
W = be repaired under warranty

a)

P(W n B1) = P(B1) P(W|B1) = 0.50*0.25 = 0.125 [answer]

b)

P(W) = P(B1) P(W|B1) + P(B2) P(W|B2) + P(B3) P(W|B3)

= 0.50*0.25 + 0.30*0.20 + 0.20*0.10

= 0.205 [answer]

c)

Thus,

P(B1|W) = P(B1) P(W|B1) / P(W) = 0.609756098
P(B2|W) = P(B2) P(W|B2) / P(W) = 0.292682927
P(B3|W) = P(B3) P(W|B3) / P(W) = 0.097560976 [answer]