More dice.....

You can calculate the EXACT probabilities for 5 dice + 6 dice + 7 dice + 8 dice + 9 dice + 10 dice as follows: [10C5 * 5^5] + [10C6 * 5^4] + [10C7 * 5^3] + [10C8 * 5^2] + [10C9 * 5^1] + [10C10 * 5^0]=

787,500 + 131,250 + 15,000 + 1,125 + 50 + 1 =934,926 / 6^10 =0.0154619667....etc.

10 fair six-sided dice are rolled. What is the probability of getting AT LEAST 5 threes?

\(\begin{array}{|rcll|} \hline P(X\ge5) &=& 1-P(X\le4) \\ &=& 1-\text{binomcdf}(10,\dfrac16,4) \\ &=& 0.01546196671 \\ \hline \end{array}\)