Probability - Gravner-30.a

Question

Roll a fair die $10$ times.

a) Compute the probability that at least one number occurs exactly $6$ times

Comments

Total sequences = $6^{10}$

Favorable outcomes = $\binom{10}{6} \left ( 6 \times 5^4 \right )$

why $\binom{10}{6} \left ( 6 \times 5^4 \right )$? because there are $\binom{10}{6}$ ways for choosing 6 places for the number which is occurring 6 times and

6 times to choose that number and remaining 4 places can be filled with 5 other numbers.

Probability = $\large \frac{\binom{10}{6} \left ( 6 \times 5^4 \right )}{6^{10}}$