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No. of balanced dice which are rolled is 7.

Total no. of possible equally likely rolls is\(\left( {{n_1}} \right)\)=\({6^7}\).

If six different numbers appear at least once, one number has to appear twice, and the other five have to appear once.

Then by Multinomial coefficient, we can find the no. of outcomes, i.e.,

\(\begin{aligned}{}{n_2} &= \frac{{7!}}{{2!{{\left( {1!} \right)}^5}}}\\ &= \frac{{7!}}{{2!}}\\ &= 2520\end{aligned}\)

For each of the other five numbers, there is an equal no. of outcomes in which each no. appears at least once.

The no. of outcomes is

\(\begin{aligned}{}{n_3} &= 6 \times {n_2}\\ &= 6 \times 2520\\ &= 15120\end{aligned}\)

The required probability is

\(\begin{aligned}{}p &= \frac{1}{{{n_1}}} \times {n_3}\\ &= \frac{1}{{{6^7}}} \times 15120\\ &= 0.05401\end{aligned}\)

Hence, the probability of each of six different numbers appear at least once is 0.05401.