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Math Help - Coin probability problem!

  1. #1
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    Coin probability problem!




    A fair coin is flipped 60 times, what is the percent chance of flipping heads 25 or fewer times?

    would you solve it this way?

    binomal
    B-(60,0.5)
    P(X< or = 25)
    so it's 60Cr * 0.5^(r) * 0.5^(60-r)
    where r=0,1,2,3,4,5,......23,24,25

    then that times 100 to get %.

    =12.25%
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  2. #2
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    Quote Originally Posted by BabyMilo View Post



    A fair coin is flipped 60 times, what is the percent chance of flipping heads 25 or fewer times?

    would you solve it this way?
    Yes,

    that would be right.

    You could also use the Normal approximation to the Binomial.

    \mu=np=30

    \sigma=\sqrt{npq}=\sqrt{15}

    Z=\frac{25-30}{\sqrt{15}}=-1.29

    Applying the correction factor \frac{0.7655}{\sqrt{np}}=0.14

    Z=-1.29+0.14=-1.15

    which corresponds to 12.5%
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  3. #3
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    Quote Originally Posted by Archie Meade View Post
    Yes,

    that would be right.

    You could also use the Normal approximation to the Binomial.

    \mu=np=30

    \sigma=\sqrt{npq}=\sqrt{15}

    Z=\frac{25-30}{\sqrt{15}}=-1.29

    Applying the correction factor \frac{0.7655}{\sqrt{np}}=0.14

    Z=-1.29+0.14=-1.15

    which corresponds to 12.5%
    no idea what you did there.
    but thanks for confirming.
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  4. #4
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    That's ok!

    It's a shortcut taken by comparing the Binomial distribution with the Normal distribution.

    The calculation can then be done more quickly (especially when we'd have to sum a substantial number of terms of the Binomial expansion).

    However, you'd need to have covered the Normal distribution for this.
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