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Math Help - Basic probablity question.

  1. #1
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    Basic probablity question.

    Three coins are tossed. Find the probabilikty that they all land heads up for each known condition.

    P(X|Y)=P(X)*P(Y)/P(Y)

    6. The first coin shows a head.
    My work: 2*2*2=8 possibilities
    8 Pr 2=56 ways to choose that so it fits that one shows a head..
    (1/56)*(1/8)/(1/8)=1/56?
    7. At least one coin shows a head.
    8. At least two coins show heads
    ...that's all I know how to work out so far.
    Also, if somebody could explain how this logically makes sense, I'd really appreciate it! (Something like a tangible example or whatnot would be great)
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  2. #2
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    Hello, puzzledwithpolynomials!

    Three coins are tossed.
    Find the probabilikty that they all land Heads up for each known condition.
    With only three coins, we can list the eight possible outcomes.

    . . \begin{array}{cccc} 1&H&H&H \\ 2&H&H&T \\ 3&H&T&H \\ 4&H&T&T \\ 5&T&H&H \\ 6&T&H&T \\ 7&T&T&H \\ 8&T&T&T \end{array}

    and we don't need any fancy formulas.




    6. The first coin shows a Head.

    The first is a Head; there are 4 cases: . HHH, HHT, HTH, HTT

    Among them, one case has all Heads.


    Therefore: . P(\text{all H }|\text{ 1st H}) \:=\:\frac{1}{4}




    7. At least one coin shows a Head.

    "At least one Head", there are 7 cases: . HHH,HHT,HTH, HTT, THH, THT, TTH

    Among them, one case has all Heads.

    Therefore: . P(\text{all H }|\text{ at least 1 H}) \:=:\frac{1}{7}




    8. At least two coins show Heads

    "At least 2 Heads", there are 4 cases: . HHH, HHT, HTH, THH

    Among them, one case has all Heads.

    Therefore: . P(\text{all H }|\text{ at least 2 H}) \:=\:\frac{1}{4}

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