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](https://tex.z-dn.net/?f=P%28%5Cfrac%7BD_%7B1%7D%20%7D%7BR%7D%20%29%20%20%3D0.6097)
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](https://tex.z-dn.net/?f=P%28%5Cfrac%7BD_%7B2%7D%20%7D%7BR%7D%20%29%20%3D0.2926)
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](https://tex.z-dn.net/?f=P%28%5Cfrac%7BD_%7B3%7D%20%7D%7BR%7D%20%29%20%3D0.09756)
Step-by-step explanation:
<u><em>Explanation</em></u>:-
Given data
<em>Let D₁ be the event of brand 1 DVD players</em>
<em>Given P( D₁) = 50% = 0.5</em>
<em>Let D₂ be the event of brand 2 DVD players</em>
<em>Given P( D₂) = 30% = 0.3</em>
<em>Let D₃ be the event of brand 3 DVD players</em>
<em>Given P(D₃ ) = 20% = 0.2</em>
<em>Let 'R' be the event of brand 1's DVD players require warranty repair work</em>
<em>Given data</em>
![P(\frac{R}{D_{1} } ) = 0.25](https://tex.z-dn.net/?f=P%28%5Cfrac%7BR%7D%7BD_%7B1%7D%20%7D%20%20%29%20%3D%20%200.25)
![P(\frac{R}{D_{2} } ) = 0.20](https://tex.z-dn.net/?f=P%28%5Cfrac%7BR%7D%7BD_%7B2%7D%20%7D%20%20%29%20%3D%20%200.20)
![P(\frac{R}{D_{3} } ) = 0.10](https://tex.z-dn.net/?f=P%28%5Cfrac%7BR%7D%7BD_%7B3%7D%20%7D%20%20%29%20%3D%20%200.10)
<em>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</em>
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<em> </em><em> </em><em>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</em>
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<em> 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</em>
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