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    Hello,
    Does anybody know how to answer this tricky Maths question? I've been stuck on this for a while now.
    Many thanks.
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    I know i can use a probability tree but i don't know where to go from there.
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    (Original post by Blaze3211)
    I know i can use a probability tree but i don't know where to go from there.
    Well the probability of throwing a head is 3/4 and a tail is 1/4
    Throwing at least two heads in three throws means either...(H,H,T) or (T,H,H) or (H,T,H) or (H,H,H) Those are all the possibilities.
    Using that info, you can work out the answer
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    To get the probability of at least 2 heads, you have to work out the probability of getting:
    All three heads and
    Head, Head, Tail (x3 as there are three variants of it)

    All three heads = 0.75 x 0.75 x 0.75 = 0.421 (3.d.p)
    HHT = 0.75 x 0.75 x 0.25 = 0.141 (3.d.p again)

    Multiply by 3 (0.141 x 3 = 0.421 again)

    Add them (0.842)
    So the probability of getting at least 2 heads would be 0.842
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    (Original post by Blaze3211)
    Hello,
    Does anybody know how to answer this tricky Maths question? I've been stuck on this for a while now.
    Many thanks.
    (Original post by homeland.lsw)
    Well the probability of throwing a head is 3/4 and a tail is 1/4
    Throwing at least two heads in three throws means either...(H,H,T) or (T,H,H) or (H,T,H) or (H,H,H) Those are all the possibilities.
    Using that info, you can work out the answer
    so multiply to find single probabilities of each possibilities then add them all up
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    (Original post by Fractite)

    Add them (0.842)
    So the probability of getting at least 2 heads would be 0.842
    Oh sorry, was I supposed to show you how to work it out and not just give you the answer?

    Oops..
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    Thanks a lot all, i got it
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    (Original post by Blaze3211)
    Thanks a lot all, i got it
    great! Anymore questions, don't hesitate to ask. The members of the Maths forum (especially Zacken ) are very helpful and prompt when answering questions!
 
 
 

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