The problem involves a binomial distribution where the number of trials is 10, and the probability of success (rolling a six) is \(\frac{1}{6}\).
We need to find the probability of getting between 3 and 5 sixes inclusive, i.e., \(P(3 \leq X \leq 5)\).
This can be calculated as:
\(P(3, 4, 5) = \binom{10}{3} \left( \frac{1}{6} \right)^3 \left( \frac{5}{6} \right)^7 + \binom{10}{4} \left( \frac{1}{6} \right)^4 \left( \frac{5}{6} \right)^6 + \binom{10}{5} \left( \frac{1}{6} \right)^5 \left( \frac{5}{6} \right)^5\)
Calculating each term:
\(\binom{10}{3} \left( \frac{1}{6} \right)^3 \left( \frac{5}{6} \right)^7 = 0.155045\)
\(\binom{10}{4} \left( \frac{1}{6} \right)^4 \left( \frac{5}{6} \right)^6 = 0.054253\)
\(\binom{10}{5} \left( \frac{1}{6} \right)^5 \left( \frac{5}{6} \right)^5 = 0.012924\)
Adding these probabilities gives:
\(0.155045 + 0.054253 + 0.012924 = 0.222\)
