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Mathematics 7 Online
OpenStudy (anonymous):

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

OpenStudy (kinggeorge):

\[{4 \over 24} \times {20 \over 24} \]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.

OpenStudy (anonymous):

But the answer should be 11/138.

OpenStudy (kinggeorge):

11/138? Give me second.

OpenStudy (kinggeorge):

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.

OpenStudy (kinggeorge):

This solution does indeed give you \[11 \over 138\]

OpenStudy (anonymous):

Thanks a lot! Can you work out ben's probability using the addition rule?

OpenStudy (kinggeorge):

What are you using as the addition rule?

OpenStudy (anonymous):

For ben: P(Car) + P(Bicycle) - P(Bicycle and car)

OpenStudy (kinggeorge):

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)=\({3 \over 23}+{8 \over 23}={11 \over 23}\)

OpenStudy (kinggeorge):

This works because Ben only gets one prize. If he were to get more than one prize, this wouldn't work.

OpenStudy (anonymous):

The addition rule is: P(A U B) = P(A) + P(B) - P(A and B).

OpenStudy (anonymous):

So how come you don't need to subtract P(A and B)?

OpenStudy (kinggeorge):

In this case P(A and B)=0 since he can only get one prize.

OpenStudy (anonymous):

Oh I see! Thanks!

OpenStudy (kinggeorge):

you're welcome.

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