Question

Suppose you are given either a fair dice or an unfair dice (6-sided). You have no basis for considering either dice more likely before you roll it and observe an outcome. For the fair dice, the chance of observing “3” is 1/6. For the unfair dice, the chance of observing “3” is 1/3. After rolling the unknown dice, you observe the outcome to be 3. What is the new probability that the die you rolled is fair

Answer 1

Answer: Our required probability is 0.83.

Step-by-step explanation:

Since we have given that

Number of dices = 2

Number of fair dice = 1

Probability of getting a fair dice P(E₁) = $\dfrac{1}{2}$

Number of unfair dice = 1

Probability of getting a unfair dice  P(E₂) = $\dfrac{1}{2}$

Probability of getting a 3 for the fair dice P(A|E₁)= $\dfrac{1}{6}$

Probability of getting a 3 for the unfair dice P(A|E₂) = $\dfrac{1}{3}$

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":

$P(E_1|A)=\dfrac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)}\\\\(E_1|A)=\dfrac{0.5\times 0.16}{0.5\times 0.16+0.5\times 0.34}\\\\P(E_1|A)=0.83$

Hence, our required probability is 0.83.