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. Each manufacturer offers one year warranty on parts and labor. It is known that 25% of brand 1 DVD players require warranty repair work whereas the corresponding percentage for brands 2 and 3 are 20% and 10% respectively. The probability that a randomly selected purchaser has a DVD player that will need repair while under warranty, is:
Step-by-step Solution:
1. Define the probabilities: \[ P(B_1) = 0.50 \] \[ P(B_2) = 0.30 \] \[ P(B_3) = 0.20 \] 2. Define the conditional probabilities: \[ P(R|B_1) = 0.25 \] \[ P(R|B_2) = 0.20 \] \[ P(R|B_3) = 0.10 \] 3. Apply the law of total probability: The total probability \( P(R) \) that a randomly selected DVD player requires repair is: \[ P(R) = P(B_1) \cdot P(R|B_1) + P(B_2) \cdot P(R|B_2) + P(B_3) \cdot P(R|B_3) \] 4. Calculate the total probability: \[ P(R) = (0.50 \times 0.25) + (0.30 \times 0.20) + (0.20 \times 0.10) \] \[ P(R) = 0.125 + 0.06 + 0.02 \] \[ P(R) = 0.205 \] Therefore, the probability that a randomly selected purchaser has a DVD player that will need repair while under warranty is: \[ \boxed{0.205} \]