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Math Help - Probability - coin tosses

  1. #1
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    Probability - coin tosses

    Need help with a probability problem that is causing me some difficulty.
    A coin is tossed 8 times. What is the probability of more than six tails?

    Thanks.

    garrett88
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  2. #2
    Kai
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    Hmm, probability to get a tail is 1/2( am assuming an unbiased coin is being used )

    On one throw it is 1/2.

    Probability to get more than 6 tails = prob to get 7 tail + prob to get 8 tail

    Hopes it helps
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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by garrett88 View Post
    Need help with a probability problem that is causing me some difficulty.
    A coin is tossed 8 times. What is the probability of more than six tails?

    Thanks.

    garrett88

    This is a binomial probibility

    The prob of 8 tails is  \left( \frac{1}{2} \right)^8
    and there is only one way this can happen.

    The prob of 7 tails is \left( \frac{1}{2} \right)^7\left( \frac{1}{2} \right)^1 but this can happen 8 ways

    \left( \frac{1}{2} \right)^8+8\left( \frac{1}{2} \right)^8=9\left( \frac{1}{2} \right)^8=\frac{9}{256}
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  4. #4
    Kai
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    Quote Originally Posted by TheEmptySet View Post
    This is a binomial probibility

    The prob of 8 tails is  \left( \frac{1}{2} \right)^8
    and there is only one way this can happen.

    The prob of 7 tails is \left( \frac{1}{2} \right)^7\left( \frac{1}{2} \right)^1 but this can happen 8 ways

    \left( \frac{1}{2} \right)^8+8\left( \frac{1}{2} \right)^8=9\left( \frac{1}{2} \right)^8=\frac{9}{256}
    I thought the oder of obtaining a head or a tail is not important

    So it would be simply (1/2)^7 * (1/2) + (1/2)^8 = 1/128
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  5. #5
    Behold, the power of SARDINES!
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    Quote Originally Posted by Kai View Post
    I thought the oder of obtaining a head or a tail is not important

    So it would be simply (1/2)^7 * (1/2) + (1/2)^8 = 1/128
    Let H be heads and T be tails If you toss a coint eight times this is the possible ways you can get 1 head and 7 tails

    {H,T,T,T,T,T,T,T}
    {T,H,T,T,T,T,T,T}
    {T,T,H,T,T,T,T,T}
    {T,T,T,H,T,T,T,T}
    {T,T,T,T,H,T,T,T}
    {T,T,T,T,T,H,T,T}
    {T,T,T,T,T,T,H,T}
    {T,T,T,T,T,T,T,H}

    There are a total of 8 ways

    You could also use the formula for binomial prob

    P(x) =\binom{8}{x}P^{x}(1-P)^{8-x}

    P(7)=\binom{8}{7} \left( \frac{1}{2} \right)^7 \left( \frac{1}{2}\right)^1=8 \cdot \left( \frac{1}{2} \right)^8

    P(8)=\binom{8}{8} \left( \frac{1}{2} \right)^8 \left( \frac{1}{2}\right)^0=\left( \frac{1}{2} \right)^8

    This gives the same answer.

    I hope this clear it up.
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  6. #6
    Bar0n janvdl's Avatar
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    Quote Originally Posted by garrett88 View Post
    Need help with a probability problem that is causing me some difficulty.
    A coin is tossed 8 times. What is the probability of more than six tails?

    Thanks.

    garrett88
    Binomial Distribution

    Work out the complement of less(or equal to) than 6.



     = 1 - [ {8 \choose 1} \left( \frac{1}{2} \right) ^{1} \left( \frac{1}{2} \right) ^{7}  + {8 \choose 2} \left( \frac{1}{2} \right) ^{2} \left( \frac{1}{2} \right) ^{6} + {8 \choose 3} \left( \frac{1}{2} \right) ^{3} \left( \frac{1}{2} \right) ^{5}  + {8 \choose 4} \left( \frac{1}{2} \right) ^{4} \left( \frac{1}{2} \right) ^{4} + {8 \choose 5} \left( \frac{1}{2} \right) ^{5} \left( \frac{1}{2} \right) ^{3} + {8 \choose 6} \left( \frac{1}{2} \right) ^{6} \left( \frac{1}{2} \right) ^{2} ]
    Last edited by janvdl; April 30th 2008 at 02:31 PM. Reason: Forgot last term
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