91Ó°ÊÓ

It is given that 5 balanced dice are rolled. So, n=5 here. Let X be the random variable denoting the outcome on the die.

It is a case of a multinomial distribution.

The pdf of a standard multinomial distribution is:

\(\)\(f\left( {x|n,p} \right) = \left\{ \begin{array}{l}\frac{{n!}}{{{x_1}! \ldots {x_n}!}}{p_1}^{{x_1}} \ldots {p_k}^{{x_k}},if\,{x_1} + \ldots + {x_k} = n\\0,otherwise\end{array} \right.\)

In this case, n=5

The probability of occurrence of 1 is denoted by \({p_1}\) , and the probability of occurrence of 4 is denoted by \({p_2}\).

\(\begin{array}{l}{p_1} = \frac{1}{6}\\{p_2} = \frac{1}{6}\\\left( {1 - {p_1} - {p_2}} \right) = \frac{4}{6}\end{array}\)

Since 1 and 4 should occur an equal number of times in a trial of 5, therefore, either they both can equally occur once or twice or zero times.

Therefore, the required probability is

\(\begin{array}{c}{p_1} = 1\,\,time\,\,and\,{p_2} = 1\,\,time\,\\ = \frac{{5!}}{{1!1!3!}}{\left( {\frac{1}{6}} \right)^1}{\left( {\frac{1}{6}} \right)^1}{\left( {\frac{4}{6}} \right)^3}\\ = \frac{{5 \times 4 \times {4^3}}}{{{6^5}}}\end{array}\)

\(\begin{array}{c}{p_1} = 2\,\,time\,\,and\,{p_2} = 2\,\,time\,\\ = \frac{{5!}}{{2!2!1!}}{\left( {\frac{1}{6}} \right)^2}{\left( {\frac{1}{6}} \right)^2}{\left( {\frac{4}{6}} \right)^1}\\ = \frac{{30 \times 4}}{{{6^5}}}\end{array}\)

\(\begin{array}{c}{p_1} = 0\,\,time\,\,and\,{p_2} = 0\,\,time\,\\ = \frac{{5!}}{{0!0!5!}}{\left( {\frac{1}{6}} \right)^0}{\left( {\frac{1}{6}} \right)^0}{\left( {\frac{4}{6}} \right)^5}\\ = \frac{{{4^5}}}{{{6^5}}}\end{array}\)

Therefore, adding them up, the required value is

\( = \frac{{2424}}{{{6^5}}}\)