Aspire Faculty ID #10439 · Topic: NIMCET 2014 · Just now
NIMCET 2014

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:

Solution

This is a Total Probability Problem.
Given Information:
$P(B_1) = 0.50, $
$ P(B_2) = 0.30, $
$P(B_3) = 0.20$

$P(R|B_1) = 0.25$
$P(R|B_2) = 0.20$
$P(R|B_3) = 0.10$


Law of Total Probability:

$P(R) = \sum_{i=1}^{3} P(R|B_i) \cdot P(B_i)$

$P(R) = P(R|B_1)\cdot P(B_1) + P(R|B_2)\cdot P(B_2) + P(R|B_3)\cdot P(B_3)$

Substituting Values:

$P(R) = (0.25)(0.50) + (0.20)(0.30) + (0.10)(0.20)$
$P(R) = 0.125 + 0.060 + 0.020$

Final Answer:

$\boxed{P(R) = 0.205}$
$\therefore $
The probability that a randomly selected purchaser will need warranty repair is $ \mathbf{20.5\%}$
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