Question

A chain of video stores sells three different brands of DVD players. Of its DVD player sales, 50% are brand 1 (the least expensive), 30% are brand 2, and 20% are brand 3. Each manufacturer offers a 1-year warranty on parts and labor. It is known that 25% of brand 1's DVD 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 DVD player that will need repair while under warranty? Ii. What is the probability that a randomly selected purchaser has a DVD player that will need repair while under warranty? Iii. If a customer returns to the store with a DVD player that needs warranty repair work, what is the probability that it is a brand 1 DVD player? A brand 2 DVD player? A brand 3 DVD player?

Answer 1

Answer:

i) If a customer returns to the store with a DVD player that needs warranty repair work, what is the probability that it is a brand 1 DVD player

$P(\frac{D_{1} }{R} ) =0.6097$

ii)  If a customer returns to the store with a DVD player that needs warranty repair work, what is the probability that it is a brand 2 DVD player

$P(\frac{D_{2} }{R} ) =0.2926$

iii) If a customer returns to the store with a DVD player that needs warranty repair work, what is the probability that it is a brand 2 DVD player

$P(\frac{D_{3} }{R} ) =0.09756$

Step-by-step explanation:

Explanation:-

Given data

Let D₁ be the event of brand 1 DVD players

Given P( D₁) = 50% = 0.5

Let D₂ be the event of brand 2 DVD players

Given P( D₂) = 30% = 0.3

Let D₃ be the event of brand 3 DVD players

Given P(D₃ ) = 20% = 0.2

Let 'R' be the event of brand 1's DVD players require warranty repair work

Given data

         $P(\frac{R}{D_{1} } ) = 0.25$

         $P(\frac{R}{D_{2} } ) = 0.20$

         $P(\frac{R}{D_{3} } ) = 0.10$

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

         $P(\frac{D_{1} }{R} ) = \frac{P(D_{1} )P(\frac{R}{D_{1} }) }{P(D_{1} )P(\frac{R}{D_{1} }) +P(D_{2}) P(\frac{R}{D_{2} } )+P(D_{3}P(\frac{R}{D_{3} } ) }$

     Substitute all values

     

       $P(\frac{D_{1} }{R} ) = \frac{0.50 X 0.25 }{0.50 X 0.25+0.30 X 0.20 + 0.20 X 0.10 }$

       $P(\frac{D_{1} }{R} ) = \frac{0.125}{0.205} =0.6097$

     

   If a customer returns to the store with a DVD player that needs warranty repair work, what is the probability that it is a brand 2 DVD player

        $P(\frac{D_{2} }{R} ) = \frac{P(D_{2} )P(\frac{R}{D_{2} }) }{P(D_{1} )P(\frac{R}{D_{1} }) +P(D_{2}) P(\frac{R}{D_{2} } )+P(D_{3}P(\frac{R}{D_{3} } ) }$

       $P(\frac{D_{2} }{R} ) = \frac{0.30 X 0.20 }{0.50 X 0.25+0.30 X 0.20 + 0.20 X 0.10 }$

      $P(\frac{D_{2} }{R} ) = \frac{0.06}{0.205} =0.2926$

If a customer returns to the store with a DVD player that needs warranty repair work, what is the probability that it is a brand 3 DVD player

        $P(\frac{D_{3} }{R} ) = \frac{P(D_{3}) P(\frac{R}{D_{3} } )}{P(D_{1} )P(\frac{R}{D_{1} }) +P(D_{2}) P(\frac{R}{D_{2} } )+P(D_{3}P(\frac{R}{D_{3} } ) }$

        $P(\frac{D_{3} }{R} ) = \frac{0.20 X 0.10 }{0.50 X 0.25+0.30 X 0.20 + 0.20 X 0.10 }$

       $P(\frac{D_{3} }{R} ) = \frac{0.02}{0.205} =0.09756$