# If a die is rolled 30 times, there are 6^{30}

If a die is rolled 30 times, there are $$\displaystyle{6}^{{{30}}}$$ different sequences possible. The following question asks how many of these sequences satisfy certain conditions.
What fraction of these sequences have exactly three 4s and three 5s?

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Laura Worden
Let 'S' be the sample space of rolling dice once.
$$\displaystyle{S}={\left\lbrace{1},{2},{3},{4},{5},{6}\right\rbrace}$$
Assuming a dice is fair, all the outcomes are equally likely.
When a dice is rolled once, there are $$\displaystyle{6}$$ possible outcomes, when a dice is rolled twice, there are $$\displaystyle{6}^{{{2}}}$$ possible outcomes, and, when the dice is rolled $$\displaystyle{n}$$ times, there are $$\displaystyle{6}^{{{n}}}$$ possible outcomes.
Here, the dice is rolled $$\displaystyle{30}$$ times. Thus, the total number of possible outcomes are $$\displaystyle{6}^{{{30}}}$$.
There are $$\displaystyle{30}$$ places available in each sequence, the number of ways of selecting exactly three $$\displaystyle{4}{S}={30}{C}^{{{3}}}$$
In the remaining $$\displaystyle{27}$$ places, the number of ways of selecting exactly three $$\displaystyle{5}{S}={27}{C}^{{{3}}}$$
In the remaining $$\displaystyle{24}$$ places, any number other than $$\displaystyle{4}$$ and $$\displaystyle{5}$$ can take place.
The number of possible ways of filling the remaining $$\displaystyle{24}$$ places $$\displaystyle={4}^{{{24}}}$$
The total number of sequences with exactly three $$\displaystyle{4}{s}$$ and three $$\displaystyle{5}{s}$$ $$\displaystyle={\left({30}{C}^{{{3}}}\right)}{\left({27}{C}^{{{3}}}\right)}{\left({4}^{{{24}}}\right)}$$
Use the formula to compute the required probability:
$$\displaystyle{R}{e}{q}{u}{i}{r}{e}{d}\ {\frac{{t}}{{i}}}{o}{n}={\frac{{{T}{o}{t}{a}{l}\ \nu{m}{b}{e}{r}\ {o}{f}\ {s}{e}{q}{u}{e}{n}{c}{e}{s}\ {w}{i}{t}{h}\ {e}{x}{a}{c}{t}{l}{y}\ {t}{h}{r}{e}{e}\ {4}{s}\ {\quad\text{and}\quad}\ {t}{h}{r}{e}{e}\ {5}{s}}}{{{T}{o}{t}{a}{l}\ \nu{m}{b}{e}{r}\ {o}{f}\ {p}{o}{s}{s}{i}{b}\le\ {o}{u}{t}{c}{o}{m}{e}{s}}}}$$
PSKRequired\ fraction=\frac{(30C^{3})(27C^{3})(4^{24})}{6^{30}}
$$\displaystyle={\frac{{{\left({4060}\right)}{\left({2925}\right)}{\left({281474976710656}\right)}}}{{{221073919720733000000000}}}}={0.01512}$$