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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?

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

The second line is indicative of what you say, its the probability of either choosing the unfair or fair die, times the respective conditional probability using that die in rolling 4 three times.

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  • $\begingroup$ What about the cases when you pick two loaded and one fair die or two fair and one loaded? $\endgroup$ – Michael Chernick Sep 14 '17 at 1:45
  • $\begingroup$ @MichaelChernick is there replacement? "You pick a die at random from the bag and roll it three times". I interpret this as you just keep rolling the one die. $\endgroup$ – Tilefish Poele Sep 14 '17 at 1:51
  • $\begingroup$ I do not believe there is replacement. $\endgroup$ – J_Heads Sep 14 '17 at 1:53
  • $\begingroup$ Sorry I assumed replacement but you are right. $\endgroup$ – Michael Chernick Sep 14 '17 at 1:57
  • $\begingroup$ no worries. I'm just trying to catch up to glen_b $\endgroup$ – Tilefish Poele Sep 14 '17 at 2:01

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