Question

Please solve this probability and statistics problem The lecturer has four 6-sided dice, of which three are regular (results 1, ...,6 are equally probable). One die is deformed so that results 1 and 6 each have probability 0.3, and each other result has probability 0.1. The lecturer has chosen one die randomly and rolled it five times, obtaining the result sequence (6, 6, 2, 4, 6). We denote = 1 if the chosen die is regular, and = 2 if it is deformed. (a) Find the probability of obtaining exactly this result sequence, if the die is regular. (b) Find the probability of obtaining exactly this result sequence, if the die is deformed. (c) Find the posterior distribution of O. (d) Using that posterior distribution, calculate the probability of obtaining 10 sixes, if the chosen die is rolled 10 more times.

Answer 1

a.

if die is regular

probability for any number = 1/6

P(6,6,2,4,6 | θ = 1) = P(6)*P(6)*P(2)*P(4)*P(6)

= (1/6)^5

= 0.0001286

b.

P(6,6,2,4,6 | θ = 2) = P(6)*P(6)*P(2)*P(4)*P(6)

= 0.3*0.3*0.1*0.1*0.3

= 0.00027

c.

P(6,6,2,4,6) = P(6,6,2,4,6 | θ = 1) * P(θ = 1) + P(6,6,2,4,6 | θ = 1) * P(θ = 2)

= 0.0001286 * (3/4) + 0.00027 *(1/4)

= 0.00016395

posterior distribution of θ :

P(θ = 1 | 6,6,2,4,6) = P(6,6,2,4,6 | θ = 1) * P(θ = 1) / P(6,6,2,4,6)

= 0.0001286 * (3/4) / 0.00016395

= 0.588289

P(θ = 2 | 6,6,2,4,6) = P(6,6,2,4,6 | θ = 2) * P(θ = 2) / P(6,6,2,4,6)

= 0.00027 *(1/4) / 0.00016395

= 0.411711

d.

p(10 sixes , θ = 1 | previous sequence : 6,6,2,4,6) = P(10 sixes | θ = 1)*P(θ = 1 | 6,6,2,4,6)

= P(six | θ = 1)^10 * P(θ = 1 | 6,6,2,4,6)

= (1/6)^10 * 0.588289

= 9.72922448*10^(-9)

p(10 sixes , θ = 2 | previous sequence : 6,6,2,4,6) = P(10 sixes | θ = 2)*P(θ = 2 | 6,6,2,4,6)

= P(six | θ = 2)^10 * P(θ = 2 | 6,6,2,4,6)

= (0.3)^10 * 0.411711

= 0.00000243111

P(10 sixes | previous sequence = 6,6,2,4,6) = p(10 sixes , θ = 1 | previous sequence : 6,6,2,4,6) + p(10 sixes , θ = 2 | previous sequence : 6,6,2,4,6)

= 9.72922448*10^(-9) + 0.00000243111

= 0.00000244083

(please UPVOTE for the hard work)