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Math Help - A card is drawn from a standard deck. Not red or ace?

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    A card is drawn from a standard deck. Not red or ace?

    A card is drawn from a standard deck of playing cards. What is the probability that it will not be a red or ace?

    My Solution (which seems to be incorrect):

    I would say, the probability that it's red is 1/2
    The probability that it's ace is 4/52. On the other hand, there are two ace cards which are Red and already counted so I would say (4-2)/52 = 2/52

    so 1 - (1/2 + 2/52) = 1 - 0.53 = 0.47

    Is this correct? How to solve this problem in the form of C(n,m)?

    Thank you
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  2. #2
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    Of the 52 cards, there are 26 that are red and 2 other aces. So that means that there are 28 that are red or an ace.

    So what's the probability of not getting red or an ace?
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    Quote Originally Posted by Prove It View Post
    Of the 52 cards, there are 26 that are red and 2 other aces. So that means that there are 28 that are red or an ace.

    So what's the probability of not getting red or an ace?
    That is 1 - (28/52) = 24/52 = 0.46 ~~~ hmmmmm
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  4. #4
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    Correct.
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