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Math Help - Probability(drawing cards)

  1. #1
    Member MathBlaster47's Avatar
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    Probability(drawing cards)

    Just making sure I have my head on straight when it comes to the whole probability thing.

    One card is drawn from a deck of 52. What is the probability that it is:
    (a): A diamond?
    (b):A King?
    (c):An Ace or King?

    Working:

    (a):There are _{52}C_1 or \frac{52!}{1!51!}=52 ways of picking a card, and there are 13 diamond cards so there are _{13}C_1 or \frac{13!}{1!12!}=13 ways of picking a diamond. So there the probability of drawing a diamond card is \frac{13}{52}=\frac{1}{4}.
    (b):There are _{52}C_1 or \frac{52!}{1!51!}=52 ways of picking a card, and there are 4 Kings so there are _4K_1 or \frac{4!}{1!3!}=4 ways to draw a king. So the probability of drawing a King is  \frac{4}{52}=\frac{1}{13}.
    (c):There are _{52}C_1 or \frac{52!}{1!51!}=52 ways of picking a card, there are 4 Aces and 4 Kings so there are 8 cards to choose from so, there are _8A_1 or \frac{8!}{1!7!}=8 ways to draw either of the cards.
    Therefore, the probability of drawing either an Ace or King is \frac{8}{52}=\frac{2}{13}

    Am I doing it right?
    Last edited by MathBlaster47; March 8th 2010 at 01:47 PM.
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by MathBlaster47 View Post
    Just making sure I have my head on straight when it comes to the whole probability thing.

    One card is drawn from a deck of 52. What is the probability that it is:
    (a): A diamond?
    (b):A King?
    (c):An Ace or King?

    Working:

    (a):There are _{52}C_1 or \frac{52!}{1!51!}=52 ways of picking a card, and there are 13 diamond cards so there are _{13}C_1 or \frac{13!}{1!12!}=13 ways of picking a diamond. So there the probability of drawing a diamond card is \frac{13}{52}=\frac{1}{4}.
    (b):There are _{52}C_1 or \frac{52!}{1!51!}=52 ways of picking a card, and there are 4 Kings so there are _4K_1 or \frac{4!}{1!3!}=4 ways to draw a king. So the probability of drawing a King is  \frac{4}{52}=\frac{1}{13}.
    (c):There are _{52}C_1 or \frac{52!}{1!51!}=52 ways of picking a card, there are 4 Aces and 4 Kings so there are 8 cards to choose from so, there are _8A_1 or \frac{8!}{1!7!}=8 ways to draw either of the cards.
    Therefore, the probability of drawing either an Ace or King is \frac{8}{52}=\frac{2}{13}

    Am I doing it right?
    Looks good to me (apart from putting K instead of C when working out the chance of a king )

    Also probability questions go here: http://www.mathhelpforum.com/math-he...s-probability/
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  3. #3
    Member MathBlaster47's Avatar
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    Quote Originally Posted by e^(i*pi) View Post
    Looks good to me (apart from putting K instead of C when working out the chance of a king )

    Also probability questions go here: http://www.mathhelpforum.com/math-he...s-probability/
    Ah.....YAY! Right answer, wrong section, got it!
    Thank you!

    P.S. Could I have a mod move the thread into the right place please?
    Last edited by MathBlaster47; March 8th 2010 at 01:49 PM. Reason: Asking for a mod to move the thread.
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  4. #4
    Member mathemagister's Avatar
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    Good job, but was putting the K instead of a C just a typo, or were you referring to the first letter of "King?" I'm just asking because I know a teacher who would take a point of for that. So, watch out!
    Just trying to help

    Mathemagister
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