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Math Help - Probability of drawing cards.

  1. #1
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    Probability of drawing cards.

    You have a 52 card deck and draw 6 cards,
    I have been trying to find the probability that the 6 cards include at least one ace.

    I think you can find the complement by first finding the probability of drawing no aces and then subtract 1 from that.
    I did find that there are 20,358,520 hands possible, if this sounds even remotely right

    This is my first time posting and I only am because my Prob. and Stat. for engineers teacher is absolutely horrible

    thanks!!
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    Quote Originally Posted by mightydog78 View Post
    You have a 52 card deck and draw 6 cards,
    I have been trying to find the probability that the 6 cards include at least one ace.

    I think you can find the complement by first finding the probability of drawing no aces and then subtract 1 from that.
    I did find that there are 20,358,520 hands possible, if this sounds even remotely right

    This is my first time posting and I only am because my Prob. and Stat. for engineers teacher is absolutely horrible

    thanks!!
    Welcome to the forum!

    I also get number of ways to draw 6 cards is \binom{52}{6} = 20358520. Number of ways to draw 6 cards such that none is an ace is \binom{48}{6}. So overall the probability is

    \dfrac{\binom{52}{6}-\binom{48}{6}}{\binom{52}{6}}

    (This is the same as 1 - P(no aces drawn).)
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    1- \frac{C^{48}_6}{C^{52}_6}

    yes, first you find the probability of the complement and substract it from 1. To ensure you won't get an ace you take the deck of cards and remove all four aces from it an then you draw your six cards, thus you have to find the number of all combinatons of six elements selected out of total 48 elements (remember deck is lacking 4 aces). U guess you get the picture.
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    sweet, that makes sense. another question asked the prob. that the 6 cards have exactly one king and i got 0.00001002, which seems crazy low but thats what I got.

    then the question asked the prob. of at least one ace and one king, would this be P(A∩B)? i feel like i should subtract something...

    thanks
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    Quote Originally Posted by mightydog78 View Post
    sweet, that makes sense. another question asked the prob. that the 6 cards have exactly one king and i got 0.00001002, which seems crazy low but thats what I got.

    then the question asked the prob. of at least one ace and one king, would this be P(A∩B)? i feel like i should subtract something...

    thanks
    For the first problem, that's a lot lower than what I get.

    There are four ways to choose the one king, and after that you must select any 5 cards from 48. That's 4 * C(48,5). Divide that by C(52,6) ways to choose total, and it's approximately 0.336.

    For the next problem, one way to think of it is: ways to get no aces but possibly kings (set A); ways to get no kings but possibly aces (set B); ways to get no kings or aces (A intersect B). We want x = |A union B| = |A| + |B| - |A intersect B|. Then it will be [C(52,6) - x] / C(52,6) as before.
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    that makes sense, i forgot the says EXACTLY on king instead of AT LEAST one king.

    but for the prob. of at least one ace and one king, I still dont understand...For x I got around 0.49 and when I plugged that into the equation you have I got 0.999.
    What am i doing wrong?
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    Quote Originally Posted by mightydog78 View Post
    that makes sense, i forgot the says EXACTLY on king instead of AT LEAST one king.

    but for the prob. of at least one ace and one king, I still dont understand...For x I got around 0.49 and when I plugged that into the equation you have I got 0.999.
    What am i doing wrong?
    x should be an integer... it's the number of ways something can happen, it's not a probability.
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