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?

[tex]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} } ) }[/tex]

Substitute all values

[tex]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 }[/tex]

[tex]P(\frac{D_{1} }{R} ) = \frac{0.125}{0.205} =0.6097[/tex]

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

[tex]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} } ) }[/tex]

[tex]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 }[/tex]

[tex]P(\frac{D_{2} }{R} ) = \frac{0.06}{0.205} =0.2926[/tex]

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

[tex]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} } ) }[/tex]

[tex]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 }[/tex]

[tex]P(\frac{D_{3} }{R} ) = \frac{0.02}{0.205} =0.09756[/tex]

francocanacari francocanacari · Answer 2 · 2021-12-02T19:36:27+01:00

1) The probability that a randomly selected purchaser has bought a brand 1 DVD player that will need repair while under warranty is 12.50%.

2) The probability that a randomly selected purchaser has a DVD player that will need repair while under warranty is 20.50%.

3) If a customer returns to the store with a DVD player that needs warranty repair work, the probability that it is a brand 1 DVD player is 60.975%, a brand 2 DVD player is 29.268%, and a brand 3 DVD player is 9.756%.

Since a chain of video stores sells three different brands of DVD players, and of its DVD player sales, 50% are brand 1, 30% are brand 2, and 20% are brand 3, and each manufacturer offers a 1-year warranty on parts and labor, and 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, to determine 1) what is the probability that a randomly selected purchaser has bought a brand 1 DVD player that will need repair while under warranty, 2) what is the probability that a randomly selected purchaser has a DVD player that will need repair while under warranty, and 3) 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 and a brand 3 DVD player, the following calculations must be performed:

1)

50 x 0.25 = X
12.50 = X

2)

(50 x 0.25) + (30 x 0.20) + (20 x 0.10) = X
12.50 + 6 + 2 = X
20.50 = X

3)

12.50 x 100 / 20.50 = X
60.975 = X
6 x 100 / 20.50 = X
29.268 = X
2 x 100 / 20.50 = X
9.756 = X

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