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
[tex]\large P(A|B)=\displaystyle\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|A^c)P(A^c)}[/tex]
Where [tex]\large A^c[/tex] is the event “the die is not fair”.
Since there are 2 dice,
[tex]\large P(A)=P(A^c)=1/2[/tex]
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
[tex]\large P(B|A)=\binom{6}{2}(1/6)^2(5/6)^4=0.2009[/tex]
and [tex]\large P(B|A^c)[/tex] 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
[tex]\large P(B|A^c)=\binom{6}{2}(0.25)^2(0.75)^4=0.2966[/tex]
hence
[tex]\large P(A|B)=\displaystyle\frac{0.2009*0.5}{0.2009*0.5+0.2966*0.5}=0.4038[/tex]
