A chain of video stores sells three different brands of DVD players. Of its DVD

Question

A chain of video stores sells three different brands of DVD players. Of its DVD player sales, 50% are brand 1, 30% are brand 2, and 20% are brand 3. It is known that 25% of brand 1’s DVD players require a repair work, whereas the corresponding percentages for brand 2 and 3 are 20% and 10%, respectively.

A. What is the probability that a randomly selected purchaser has brought a brand 1 DVD player that will need a repair work?

B. What is the probability that a randomly selected purchaser has a DVD player that will need a repair work?

C. If a customer returns to the store with a DVD player that needs repair work, what is the probability that it is a brand 1 DVD player?

Explanation / Answer

Probability of DVD player is from brand 1 = 0.5 or 50%

Probability of DVD player is from brand 2 = 0.3 or 30%

Probability of DVD player is from brand 3 = 0.2 or 20%

Probability of DVD player from brand 1 require repair = 0.25 or 25%

Probability of DVD player from brand 2 require repair = 0.20 or 20%

Probability of DVD player from brand 3 require repair = 0.10 or 10%

A)

probability that a randomly selected purchaser has brought a brand 1 DVD player that will need a repair work = Probability of DVD player is from brand 1 * Probability of DVD player from brand 1 require repair = 0.5*0.25 = 0.125

B)

probability that a randomly selected purchaser has a DVD player that will need a repair work =

(Probability of DVD player is from brand 1)*(Probability of DVD player from brand 1 require repair) + (Probability of DVD player is from brand 2)*(Probability of DVD player from brand 2 require repair) +(Probability of DVD player is from brand 3)*(Probability of DVD player from brand 3 require repair)

= 0.5*0.25+ 0.3*0.20 +0.2* 0.10 = 0.205

c)

we know that P(A/B) = P(A n B)/P(B)

probability that DVD player is a brand 1 DVD player given that the customer returns to the store with a DVD player that needs repair work = probability that a randomly selected purchaser has brought a brand 1 DVD player that will need a repair work/probability that a randomly selected purchaser has a DVD player that will need a repair work

= 0.125/0.205

= 0.6098

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