Question

Assume that you have two dice, one of which is fair, and the other is biased toward landing on six, so that 0.25 of the time it lands on six, and 0.15 of the time it lands on each of 1, 2, 3, 4 and 5. You choose a die at random, and roll it six times, getting the values 4, 3, 6, 6, 5, 5. What is the probability that the die you chose is the fair die? The outcomes of the rolls are mutually independent.

Answer 1

Answer:

0.4038

Step-by-step explanation:

Let A and B be the events

A: “The die is fair”

B: “The die lands on 6 two times out of 6”

We want to determine if the die is fair given that it landed on 6 two times out of 6 tosses, that is P(A | B).

By the Bayes' theorem  

$\large P(A|B)=\displaystyle\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|A^c)P(A^c)}$

Where $\large A^c$ is the event “the die is not fair”.

Since there are 2 dice,

$\large P(A)=P(A^c)=1/2$

If the die is fair P(B | A) is the probability of getting exactly two six in a binomial experiment with probability of “success” (land on 6) 1/6 and six repeated trials  

$\large P(B|A)=\binom{6}{2}(1/6)^2(5/6)^4=0.2009$

and $\large P(B|A^c)$ is the probability of getting exactly two six in a binomial experiment with probability of “success” (land on 6) 0.25 and six repeated trials  

$\large P(B|A^c)=\binom{6}{2}(0.25)^2(0.75)^4=0.2966$

hence

$\large P(A|B)=\displaystyle\frac{0.2009*0.5}{0.2009*0.5+0.2966*0.5}=0.4038$