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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

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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.
But the answer should be 11/138.
11/138? Give me second.

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Other answers:

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.
This solution does indeed give you \[11 \over 138\]
Thanks a lot! Can you work out ben's probability using the addition rule?
What are you using as the addition rule?
For ben: P(Car) + P(Bicycle) - P(Bicycle and car)
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)=\({3 \over 23}+{8 \over 23}={11 \over 23}\)
This works because Ben only gets one prize. If he were to get more than one prize, this wouldn't work.
The addition rule is: P(A U B) = P(A) + P(B) - P(A and B).
So how come you don't need to subtract P(A and B)?
In this case P(A and B)=0 since he can only get one prize.
Oh I see! Thanks!
you're welcome.

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