180 throws with a normal dice

**sf1903** · August 7th 2009, 05:17 AM

X: numbers of 6's after 180 throws with a normal dice.
P(X>=38)=?

**Failure** · August 7th 2009, 08:18 AM

Originally Posted by sf1903

X: numbers of 6's after 180 throws with a normal dice.
P(X>=38)=?

$\begin{array}{lcl} \mathrm{P}(38\leq X) &=& 1-\mathrm{P}(X\leq 37)\\ &=& \displaystyle 1-\sum\limits_{x=0}^{37}\binom{180}{x}\cdot \left(\frac{1}{6}\right)^x\left(1-\frac{5}{6}\right)^{180-x}\\ &\approx & 0.0699673478 \end{array}$

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) $

**Zarathustra** · August 7th 2009, 08:22 AM

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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