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}}

\(\displaystyle={\frac{{{\left({4060}\right)}{\left({2925}\right)}{\left({281474976710656}\right)}}}{{{221073919720733000000000}}}}={0.01512}\)

\(\displaystyle{S}={\left\lbrace{1},{2},{3},{4},{5},{6}\right\rbrace}\)

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}}

\(\displaystyle={\frac{{{\left({4060}\right)}{\left({2925}\right)}{\left({281474976710656}\right)}}}{{{221073919720733000000000}}}}={0.01512}\)