Answer: Our required probability is 0.83.
Explanation:
Since we have given that
Number of dices = 2
Number of fair dice = 1
Probability of getting a fair dice P(E₁) =
![(1)/(2)](https://img.qammunity.org/2020/formulas/mathematics/high-school/g8uu853hd4xgpf51yzryheaugm47qujkf6.png)
Number of unfair dice = 1
Probability of getting a unfair dice P(E₂) =
![(1)/(2)](https://img.qammunity.org/2020/formulas/mathematics/high-school/g8uu853hd4xgpf51yzryheaugm47qujkf6.png)
Probability of getting a 3 for the fair dice P(A|E₁)=
![(1)/(6)](https://img.qammunity.org/2020/formulas/mathematics/high-school/hpryu5gz6b2hrulak2xp6643a8fps5q09c.png)
Probability of getting a 3 for the unfair dice P(A|E₂) =
![(1)/(3)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/jpd6zv0h9zkyvgttvfpmhqo3uswr1jq775.png)
So, we need to find the probability that the die he rolled is fair given that the outcome is 3.
So, we will use "Bayes theorem":
![P(E_1|A)=(P(E_1)P(A|E_1))/(P(E_1)P(A|E_1)+P(E_2)P(A|E_2))\\\\(E_1|A)=(0.5* 0.16)/(0.5* 0.16+0.5* 0.34)\\\\P(E_1|A)=0.83](https://img.qammunity.org/2020/formulas/mathematics/college/o6kvqj5xed5n19szvjk0gfhfx2g8izfc8x.png)
Hence, our required probability is 0.83.