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

i)

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

ii)

By Bayes\' Rule,

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]

iii)

By Bayes\' Rule,

P(B1|W) = P(B1) P(W|B1) / P(W) = 0.125 / 0.205 = 0.609756098 [answer]

P(B2|W) = P(B2) P(W|B2) / P(W) = 0.30*0.20 / 0.205 = 0.292682927 [answer]

P(B3|W) = P(B3) P(W|B3) / P(W) = 0.20*0.10 / 0.205 = 0.097560976 [answer]