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Math Help - probability/combination flipping a coin.

  1. #1
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    probability/combination flipping a coin.

    question is,
    If you flip a coin 5 times what is the probability of getting 4 heads, and 1 tails?

    total number of outcomes is 2^5 because you flip a coin there are to possible outcomes 5 times.

    Since there is no order it is a combination *this is where I am confused* I thought when you have two different things like,

    "how many 4 card hands can be formed with 2 hearts and 2 diamonds" you do 13c2 x 13c2 because you have 13 of each and you are choosing two each.

    but what I've seen searching online is when you flip a coin dice 5 times and want 4 heads it's 5c4=5? but what about the tails? shouldn't it be 5c4 x 5c1 because there are 5 heads and 5 tails? Also if this is the same with dice I'd like to know how to do that too.

    thanks.
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    Quote Originally Posted by brentwoodbc View Post
    question is,
    If you flip a coin 5 times what is the probability of getting 4 heads, and 1 tails?
    total number of outcomes is 2^5 because you flip a coin there are to possible outcomes 5 times.
    There are five ways to have 4 heads and one tail.
    THHHH
    HTHHH
    HHTHH
    HHHTH
    HHHHT
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    Quote Originally Posted by Plato View Post
    There are five ways to have 4 heads and one tail.
    THHHH
    HTHHH
    HHTHH
    HHHTH
    HHHHT
    how would you do that if there where more, like 100 head's 125 tails.
    I thought it would be 225C100 times 225C125 ?
    but it seems to be 225C100?

    Is dice the same thing? I dont see how this is different from having say 26 red cards and 5 black cards where you want a 6 card hand, it's 26c4 times 5c2
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    Quote Originally Posted by brentwoodbc View Post
    how would you do that if there where more, like 100 head's 125 tails.
    The probability of 100 heads and 125 tails is \binom{225}{100}\left(\frac{1}{2}\right)^{225}.
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    Quote Originally Posted by Plato View Post
    The probability of 100 heads and 125 tails is \binom{225}{100}\left(\frac{1}{2}\right)^{225}.
    hmm. so 225C100? I thought you had to account for the heads and tails, I guess not.
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    or maybe I'm thinking combination but it's actually a permutation.
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    Quote Originally Posted by brentwoodbc View Post
    hmm. so 225C100? I thought you had to account for the heads and tails, I guess not.
    I did account for both.
    We have 225 places. In 100 of them there will be a head.
    Of course in the others there will be a tail.
    Remember that \binom{225}{100}=\binom{225}{125}.
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    Quote Originally Posted by Plato View Post
    I did account for both.
    We have 225 places. In 100 of them there will be a head.
    Of course in the others there will be a tail.
    Remember that \binom{225}{100}=\binom{225}{125}.
    I'm just starting probability, what does this notation mean \binom{225}{100}=\binom{225}{125} permutation?
    Last edited by brentwoodbc; May 28th 2009 at 01:00 PM.
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    Quote Originally Posted by brentwoodbc View Post
    I'm just starting probability, what does \binom{225}{100}=\binom{225}{125} mean? permutation?
    No, not permutations but combinations.
    From N items choose K, a combination of N things taken K at a time.
    \binom{N}{K}=\binom{N}{N-K}=\frac{N!}{(K!)(N-K)!}
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    Quote Originally Posted by brentwoodbc View Post
    a deck has 52 cards, 12 face cards and 4 aces are removed, from these 16 cards 4 are chosen, how many combinations are possible that have at least 2 black cards.

    The answer is 1302




    Quote Originally Posted by Soroban View Post
    Hello, brentwoodbc!

    You're forgetting about the red cards . . .



    To get "at least 2 Blacks", we want: .(2 Black, 2 Red) or (3 Black, 1 Red) or (4 Black).

    . . There are: . (_8C_2)(_8C_2) = 784 ways to get 2 Black, 2 Red.

    . . There are: . (_8C_3)(_8C_1) = 448 ways to get 3 Black, 1 Red.

    . . There are: . _8C_4 = 70 ways to get 4 Black.


    Therefore, there are: . 784 + 448 + 70 \:=\:1302 ways to get at least 2 Blacks.

    This was another question from a while ago, it is whats confusing me, why di you multiply here, but not with the dice. I mean whats the difference between a red card or a black card - vs. - a heads or a tails?

    or 4 black cards and 1 red vs. 4 tails and 1 heads.
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  11. #11
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    Quote Originally Posted by brentwoodbc View Post
    This was another question from a while ago, it is whats confusing me, why di you multiply here, but not with the dice. I mean whats the difference between a red card or a black card - vs. - a heads or a tails? or 4 black cards and 1 red vs. 4 tails and 1 heads.
    The two ideas are entirely different: one, the cards, is a dependent set of acts; the other, coins, is an independent set of acts.

    By that we mean: as we deal cards the second card depends upon what the first card was.
    But in flipping a coin the second coins is not effected by what happens on the first coin.

    If we have an urn of ten red balls, five blue balls and one gold ball.
    Drawing out four balls one at a time without replacement is a dependent act.

    Drawing out four balls one at a time with replacement is an independent act.
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