Die contains six possible outcomes \(\displaystyle{\left[{1},{2},{3},{4},{5},{6}\right]}\). The number of possible outcomes when a die is rolled 35 times is \(\displaystyle{6}^{{{35}}}\)

There would be 35 slots when it is rolled 35 times. The sequence must consist of five 1's

This can be done in \(\displaystyle{35}{C}_{{5}}\) ways. After filling of five slots with 1's there would be 30 remaining slots. They can be filled with numbers \(\displaystyle{2},{3},{4},{5},{6}\) in \(\displaystyle{5}^{{{30}}}\) ways

The fraction of \(\displaystyle{6}^{{{35}}}\) sequences that have exactlyfive 1's is:

\(\displaystyle{P}{r}{o}{b}{a}{b}{i}{l}{i}{t}{y}={\frac{{{35}{C}_{{5}}\times{5}^{{{30}}}}}{{{6}^{{{35}}}}}}\)

\(\displaystyle={0.1759}\)

There would be 35 slots when it is rolled 35 times. The sequence must consist of five 1's

This can be done in \(\displaystyle{35}{C}_{{5}}\) ways. After filling of five slots with 1's there would be 30 remaining slots. They can be filled with numbers \(\displaystyle{2},{3},{4},{5},{6}\) in \(\displaystyle{5}^{{{30}}}\) ways

The fraction of \(\displaystyle{6}^{{{35}}}\) sequences that have exactlyfive 1's is:

\(\displaystyle{P}{r}{o}{b}{a}{b}{i}{l}{i}{t}{y}={\frac{{{35}{C}_{{5}}\times{5}^{{{30}}}}}{{{6}^{{{35}}}}}}\)

\(\displaystyle={0.1759}\)