A bag contains 8 fair dice as well as two rigged dice with 4 dots on all six sides

A bag contains 8 fair dice as well as two rigged dice with 4 dots on all six sides. You pick a die at random from the bag and roll it three times. What is the probability that all three rolls produce the 4-dot outcome?

My first thought process was $((1/5 * 1) + (4/5 * 1/6))^3$ due to the two cases: either you choose a unfair or fair die and add this probability. Then take the cube of the probability of rolling a 4 for a single die. Where am I going wrong?

Your thinking is right but your algebra doesn't match your statement. $((1/5 * 1) + (4/5 * 1/6))^3 \neq 1/5 * (1)^3 + 4/5*(1/6)^3$