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- Apr 14, 2013

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Tim rolls $20$ times a fair dice.

- Calculate the probability to get a "$3$" exactly $8$ times.
- The probability to get at least $k$ times a "$5$" is equal to $10,18\%$ Calculate $k$.
- Explain the relation to the term $\displaystyle{\sum_{i=0}^2B\left (10;\frac{5}{6};i\right )}$.

I have done the following:

- The probability to get a "$3$" is equal to $\frac{1}{6}$. If we throw $20$ times the dice the probability to get exactly eight "$3$" is equal to $\binom{20}{8}\cdot \left (\frac{1}{6}\right )^8\cdot \left (1-\frac{1}{6}\right )^{20-8}=\binom{20}{8}\cdot \left (\frac{1}{6}\right )^8\cdot \left (\frac{5}{6}\right )^{12}\approx 0.84\%$.

Is this correct?

$$ $$ - We have that \begin{align*}P(X\geq k)=10,18\% &\Rightarrow 1-P(X\leq k)=10,18\% \\ & \Rightarrow \sum_{i=0}^k\binom{20}{i}\cdot \left (\frac{1}{6}\right )^i\cdot \left (1-\frac{1}{6}\right )^{20-i}=89.82\% \\ & \Rightarrow \sum_{i=0}^k\binom{20}{i}\cdot \left (\frac{1}{6}\right )^i\cdot \left (\frac{5}{6}\right )^{20-i}=89.82\% \end{align*} How can we continue to get the value of $k$ ?

$$ $$ - Is this the formula we used in the previous question, just with other numbers?