Ben rolls four fair 10-sided dice. What is the probability that exactly three of the dice show a prime number?
+486 Out of a $10$-sided die, there are $4$ prime numbers: $2,3,5,7$
The proability of rolling a prime number is $\frac{4}{10}=\frac{2}{5}$.
So the probability of rolling 3 prime numbers is $(\frac{2}{5})^{3}=\frac{8}{125}$.
But wait! The problem says EXACTLY three, so we can't roll a prime number on the fourth roll.
So the probability of rolling EXACTLY 3 prime numbers is $\frac{8}{125}*\frac{6}{10}=\frac{8}{125}*\frac{3}{5}=\frac{24}{625}$.
But wait! There are ${4 \choose 3}=4$ ways to choose 3 of the four dice to roll prime numbers on, so the final probability is $\frac{24}{625}*4=\boxed{\frac{96}{625}}$
