At Least One Six in Three Rolls

It is easier to work with the complementary event of never rolling a 6.

Multiply the probability of avoiding a 6 on each independent roll.

1. 1. Complement probability

   Each roll avoids a 6 with probability $5/6$, and the rolls are independent.

   $$P(\text{no six})={\left(\frac{5}{6}\right)}^3=\frac{125}{216}$$

2. 2. Subtract from one

   The probability of at least one six is the complement of never rolling one.

   $$P=1-\frac{125}{216}=\frac{91}{216}$$

Answer: 91/216