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Math Help - Bridge game distribution of cards

  1. #1
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    Sep 2009
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    Thumbs down Bridge game distribution of cards

    63. you are dealt a 13-card bridge hand from a 52-card bridge deck.
    b. what is the probability you will be dealt at least 2 hearts?

    so i think i figured this out... just not sure if i'm missing more to the equation...

    N=52 (population)
    n=13(sample size)
    x=2(number of successes)
    number of failures = n-x
    = 13-2
    = 11
    P(at least 2 hearts) = 13 39
    2 11
    ---------
    52
    13
    = 78(1,675,986,051) / 6.53 x 10^11
    = 1.307 x 10^11 / 6.35 x 10^11
    = 0.20583
    therefore, the probability you will receive at least 2 hearts is 0.20583 or 20.583% chance.

    c. What is the probability that you will receive 5 hears, 4 clubs, 2 diamonds and 1 spade?

    so i already calculated out for 5 hearts in part a. and came to 0.12472

    N=52 still?
    n=8 (we already took 5 cards)
    x= 4
    number of failures = n-x
    = 8-4
    = 6


    P(4 clubs) = 8 39
    4 6
    ---------
    52
    8
    = 70(3,262,623)/752,538,150
    = 228,383,610/752,538,150
    = 0.30345

    now N=52
    n = 4
    x = 2
    number of failures = n-x
    = 4-2
    = 2
    P(2 diamonds) = 4 39
    2 2
    ---------
    52
    4
    = 6(741)/270,725
    = 4,446/ 270,725
    = 0.01642
    N=52
    n=2
    x=1
    number of failures=n-x
    = 2-1
    = 1
    P(1 club) = 2 39
    1 1
    ---------
    52
    2
    = 2(39)/1,326
    = 78/1,326
    = 0.05882

    then to find the total probability it would be;

    Pt = P(5 hearts) + P(4 spades) + P(2 diamonds) + P(1 club)
    = 0.12472 + 0.30345 + 0.01642 + 0.05882
    = 0.50341

    Therefore the probability that you will be dealt 5 hearts, 4 spades, 2 diamonds and 1 club is 0.50341 or 50.341%?

    Again with this question i just want to make sure that i've got this figured out the right way...
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  2. #2
    Member
    Joined
    Mar 2009
    Posts
    133
    I cant really understand the working you have used for (b) but this is how I have done it;
    I think it is right..
    P(\mbox{2 or more hearts})=1-[P(\mbox{0 hearts})+P(\mbox{1 heart})]
     <br />
P(\mbox{0 hearts})=\frac{13C0\cdot 39C13}{52C13}=\frac{ \frac{13!}{0!13!}\cdot \frac{39!}{13!\cdot 26!} } {\frac {52!}{13! \cdot 39!}}=0.012791

    <br />
P(\mbox{1 hearts})=\frac{13C1\cdot 38C12}{52C13}=\frac{ \frac{13!}{1!12!}\cdot \frac{38!}{12!\cdot 26!} } {\frac {52!}{13! \cdot 39!}}=0.055427
    so
    P(\mbox{2 or more hearts})=1-(0.012791+0.055427)=0.93178161
    So 93% chance of 2 or more hearts.

    A similiar method can be used for part (c), however, you have only listed 12 cards being delt?
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