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adnanchowdhury

In a lottery there are 24 prizes allocated at random to 24 prize-winners. Ann, Ben, and Cal are three of the winners. Of the prizes, 4 are cars, 8 are bicycles and 12 are watches. Find the probability that Ann gets a car and Ben gets a car or bicycle

  • 2 years ago
  • 2 years ago

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  1. KingGeorge
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    \[{4 \over 24} \times {20 \over 24} \]The 4/24 is Ann's chance of getting a car, and the 20/24 is Ben's chance of getting a car or a bike.

    • 2 years ago
  2. adnanchowdhury
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    But the answer should be 11/138.

    • 2 years ago
  3. KingGeorge
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    11/138? Give me second.

    • 2 years ago
  4. KingGeorge
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    My bad, I was calculating the probability that Ben got a bike or a watch, not a bike or a car. In this case, the probability is given by \[{4 \over 24} \times {{3+8} \over 23}\]Where 4/24 is the probability that Ann gets a car, and the other term is the chance Ben got a car plus the chance he got a bike.

    • 2 years ago
  5. KingGeorge
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    This solution does indeed give you \[11 \over 138\]

    • 2 years ago
  6. adnanchowdhury
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    Thanks a lot! Can you work out ben's probability using the addition rule?

    • 2 years ago
  7. KingGeorge
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    What are you using as the addition rule?

    • 2 years ago
  8. adnanchowdhury
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    For ben: P(Car) + P(Bicycle) - P(Bicycle and car)

    • 2 years ago
  9. KingGeorge
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    nm, I realized I knew it. By the addition rule, Ben's probability is P(Car)=\(3 \over 23\) and P(Bike) =\(8 \over 23\) so P(Car)+P(Bike)=P(Car or Bike)=\({3 \over 23}+{8 \over 23}={11 \over 23}\)

    • 2 years ago
  10. KingGeorge
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    This works because Ben only gets one prize. If he were to get more than one prize, this wouldn't work.

    • 2 years ago
  11. adnanchowdhury
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    The addition rule is: P(A U B) = P(A) + P(B) - P(A and B).

    • 2 years ago
  12. adnanchowdhury
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    So how come you don't need to subtract P(A and B)?

    • 2 years ago
  13. KingGeorge
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    In this case P(A and B)=0 since he can only get one prize.

    • 2 years ago
  14. adnanchowdhury
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    Oh I see! Thanks!

    • 2 years ago
  15. KingGeorge
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    you're welcome.

    • 2 years ago
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