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Need a stats wiz… Simple question…

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  • #46470

    Anderson
    Participant

    How do I figure out the probablitiy of getting at least (1) heads when flipping a coin (3) times? (given that the prob. of getting a heads on one flip is 50%)
    I can’t seem to remember how to do this…
    Thanks,
    -Mike
     

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    #153627

    anonimus
    Participant

    the events could be
    H H H;  HHC; HCC ; CCC.   3 opportunity on 4
     the probability you can get at least one Head is = .75
     
     

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    #153629

    Cone
    Participant

    This is a binomial problem and your answer is wrong because you did not account for the number of ways some thing can happen
    0 heads can only happen one way and the probability is 0.125
    1 head can happen three different ways (HTT,THT, or TTH) and the probability of each is 0.125 for a total of 0.375
    2 heads can happen three different ways (HHT,HTH, or THH) and the probability of each is 0.375
    3 heads can only happen one way and the probability is 0.125.
    So the probability of 1 or more heads = 1- probability of 0 heads = 0.875.

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    #153633

    Bower Chiel
    Participant

    As with many probability probability problems this one may be solved quickly by looking at the complementary event to the one that you are interested in.P(At least one head)
    =1-P(No heads)
    =1-P(All three coins show tails)
    =1-0.5*0.5*0.5
    =0.875Sorry anonimus but I think you should have listed: –
    HHH HHT HTH THH HTT THT TTH TTT.Bower Chiel

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    #153636

    jediblackbelt
    Participant

    Maybe I am reading his question wrong, but I read it as if I flip a coin 3 times what is the probability that only one of them is heads.  If that is the case then wouldn’t you use the binomial probability? 
    I come up with a .375 chance that flipping a coin 3 times that only 1 of the coin flips will be heads. 
    Am I off base?

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    #153640

    Cone
    Participant

    Look at the words “at least” in the original problem.

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    #153643

    BritW
    Participant

    If it matters at this point – Gary is right.  Probability of filpping at least 1 head = 1-P(flipping 0 heads) = 1-.125 = .875.
    =1-{(3!/(0!(3-0)!)) x .5(to 0 power) x .5(to 3-0 power)}
    =1- {(1) x (1) x .125}
    = .875

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    #153649

    Anderson
    Participant

    Thanks for the replies.  I used the 1-p approach, but when I changed the probability of getting a heads to 20% (instead of 50%) my probability of getting at least one heads on three flips went up… which doesn’t make sense.
    (.5 x .5 x .5) = .125 = 1-p = 1-.125 = 87.5%
    (.2 x .2 x .2) = .008 = 1-.008 = 99.2%
    What am I doing wrong?
     

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    #153650

    Cone
    Participant

    It’s 1 – probability of three tails which is 1 – .8*.8*.8 or .488.

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    #153651

    BTDT
    Participant

    Mikea:In your second calculation you should use three p(tails) and subract from unity. You used three p(heads).Cheers, BTDT

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    #153652

    Anderson
    Participant

    Doh. Ok.. I feel like an idiot… not the first time either. Thanks all. =o)

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    #153663

    jediblackbelt
    Participant

    Cool.  Not off base, just misread.  Good thing this isn’t a stats test.  Thanks for the clarification.
     

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    #153670

    goodie
    Participant

    MikeA,
    the diagram would be:
     
    1st flip                      H                                        T
    2nd flip            H              T                          H             T
    3rd flip       H       T       H       T               H      T      H     T  
                  HHH HHT HTH HTT           THH THT  TTH  TTT
    combination with at least 1 H is 7 out of 8, so
    the probability of having at least 1 Head is 7/8 = .875
     
    Goodie

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