outcome from fair and non-fair dice, dice are chosen from other dice part1

Question

There're 3 dice, two are fair, one is not, it has an occurrences of $5$ instead of $4$. That is the non-fair dice is ${1,2,3,5,5,6}.$ We choose one dice from the three randomly, and throw it twice. What is the probability that two rolls will show $5$.

If the dice is fair then the probability that $5$ is rolled twice in a throw is $\frac{1}{6} \times \frac{1}{6}$, if the dice is not fair then the probability is $\frac{2}{6} \times \frac{2}{6} = \frac{1}{9}$.

How do I take into account that a dice can be chosen from three?

Answer 1

Split this up into two cases:

Case 1: Fair die

• The probability of getting a fair die is $\frac 2 3$
• The probability of two fives with a fair die is, as you found, $\frac 1 {36}$

Case 1: Non-fair die

• The probability of getting a fair die is $\frac 1 3$
• The probability of two fives with a fair die is, as you found, $\frac 1 {9}$

Then, multiply the probabilities within each case and add the results: $$\frac 2 3\cdot\frac 1 {36}+\frac 1 3\cdot\frac 1 9=\frac 1 {18}$$

Splitting events up into exhaustive cases and multiplying within the cases/adding between the cases is called the Law of Total Probability.

Answer 2

The probability to get a fair die is $\frac{2}{3}$ and the chance to get the unfair die is $\frac{1}{3}$. Combining this with your results yields $$\frac{2}{3} \cdot \frac{1}{36} + \frac{1}{3} \frac{1}{9} = \frac{1}{18}$$ as probability of throwing two fives.