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  1. #1
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    Question

    A car dealer has 8 Red, 11 Gray, and 9 Blue cars in stock Ten cars are randomly chosen to be displayed in from of the dealership. Find the probability that 4 are red and 6 are blue; 3 are red, 3 are blue and 4 are gray; 5 are gray and non are blue; all 10 are gray.
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  2. #2
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    Quote Originally Posted by batman123
    A car dealer has 8 Red, 11 Gray, and 9 Blue cars in stock Ten cars are randomly chosen to be displayed in from of the dealership. Find the probability that 4 are red and 6 are blue; 3 are red, 3 are blue and 4 are gray; 5 are gray and non are blue; all 10 are gray.
    Remember that probability is favorable outcomes divided by possible outcomes.

    1)4 Red and 6 Blue: There are 8 red he is chosing 4 thus there are _8C_4=70 different ways. There are 9 blue he is chosing 6 thus there are _9C_6=_9C_3=84 different ways. Thus there are a complete total of 70\times 84=5880 ways of chosing these types of cars. This is the favorable outcomes, now we find the possible outcomes. There are 28 cars altogether and he is taking 10, thus there are _{28}C_{10}=13123110 different ways. Thus the probability is the ratio thus,
    p=\frac{5880}{13123110}.

    Everything else is solved similarly.
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