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Math Help - 180 throws with a normal dice

  1. #1
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    Wink 180 throws with a normal dice

    X: numbers of 6's after 180 throws with a normal dice.
    P(X>=38)=?
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by sf1903 View Post
    X: numbers of 6's after 180 throws with a normal dice.
    P(X>=38)=?
    \begin{array}{lcl}<br />
\mathrm{P}(38\leq X) &=& 1-\mathrm{P}(X\leq 37)\\<br />
&=& \displaystyle 1-\sum\limits_{x=0}^{37}\binom{180}{x}\cdot \left(\frac{1}{6}\right)^x\left(1-\frac{5}{6}\right)^{180-x}\\<br />
&\approx & 0.0699673478<br />
\end{array}

    Or, you could use normal approximation to determine \mathrm{P}(X\leq 37). Because \mu_X=180\cdot \frac{1}{6}=30 and \sigma_X=\sqrt{180\cdot\frac{1}{6}\cdot \left(1-\frac{5}{6}\right)}=5 you get (by the central limit theorem)

    \mathrm{P}(38\leq X) = 1-\mathrm{P}(X\leq 37)= 1-\mathrm{P}\left(\frac{X-30}{5}\leq \frac{37-30+0.5}{5}\right) \approx 1-\Phi\left(\frac{7.5}{5}\right)<br />
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  3. #3
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    And just in case the above answer looks like gibberish, you might want to read about the binomial distribution on wikipedia:
    Binomial distribution - Wikipedia, the free encyclopedia
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