Question

Attention: I had earlier asked the same but the solution was not clear,Kindly show all the workings explicitly.

1) Your friend grabs a die at random from a drawer containing two 6 -sided dice,one 8-sided,and one 12-sided die.

She rolls the die once and reports that the result is 7.

a) Make a discrete Bayes table showing the prior ,likelihood,and posterior for the type of die rolled given the data.

b) What are your posterior odds that the die has 12 sides?

c) Given the data of the first roll,what is your probability that the next roll will be a 7?

Answer 1

Solution

Preparatory work

Let S, E and T represent the events that the die picked from the drawer is 6-sided, 8-sided and 12-sided respectively.

Let R represent the event that the result of rolling the selected die is 7.

Since there are 4 dice in the drawer, of which 2 are 6-sided, P(S) = 2/4 = ½ ……………….(1)

Similarly, P(E) = ¼ …………………...........…………………………………..………………...(2)

And P(T) = ¼ ………………………………….............…………………..……………………...(3)

Since rolling of a 6-sided die can result in any one of the six faces, 1 to 6, and 7 is not one of these, P(R/S) = 0…………………………………………………………………........……… .. (4)

Since rolling of a 8-sided die can result in any one of the eight faces, 1 to 8, P(R/E) = 1/8 ..(5)

Similarly, P(R/T) = 1/12 ……………………..........………………………..…..……………….....(6)

Back-up Theory

Conditional Probability and Bayes Theorem

If A and B are two events such that probability of B is influenced by occurrence or otherwise of A, then

Conditional Probability of B given A, denoted by P(B/A) = P(B ∩ A)/P(A)………….….(7)

P(B) = {P(B/A) x P(A)} + {P(B/A^C) x P(A^C)}…………………………………………...….(8)

If A is made up of k mutually and collectively exhaustive sub-events, A₁, A₂,..A_k,

P(B) = sum over i = 1 to k of {P(B/A_i) x P(A_i)} …………………………………………...(9)

P(A/B) = P(B/A) x {P(A)/P(B)} [Bayes Theorem]……………………..………………….(10)

Now, to work out the answer,

Connecting the defined terminology and the terminology under Back-up Theory,

S, E and T are same as A₁, A₂,A₃ and R is the same as B.

So, vide (9),

P(R) = {P(R/S) x P(S)} + {P(R/E) x P(E)} + {P(R/T) x P(T)}

= {0 x (½)} + {(1/8) x (¼)}+ {(1/12) x (¼)} [(4) to (6)]

= 5/96 ……………………………………………………………………………………..(11)

Further, vide (10),

P(S/R) ={P(R/S)P(S)}/P(R) = {0 x (½)}/(5/96) = 0 [vide (11)] …………………………(12)

P(E/R) ={P(R/E)P(E)}/P(R) = {(1/8) x (¼)}/(5/96) = 3/5 [vide (11)] ……………………(13)

P(T/R) ={P(R/T)P(T)}/P(R) = {(1/12) x (¼)}/(5/96) = 2/5 [vide (11)] ………………......(14)

Part (a)

Prior likely-hood for 6-sided die = ½ [vide (1)] ANSWER 1

Prior likely-hood for 8-sided die = ¼ [vide (2)] ANSWER 2

Prior likely-hood for 12-sided die = ¼ [vide (3)] ANSWER 3

Posterior for 6-sided die given the result is 7 is 0 [vide (12)] ANSWER 4

Posterior for 8-sided die given the result is 7 is 3/5 [vide (13)] ANSWER 5

Posterior for 8-sided die given the result is 7 is 2/5 [vide (14)] ANSWER 5

Part (b)

Posterior odds that the die has 12 sides = 2/5 [vide (14)] ANSWER 6

Part (c)

Irrespective of the data of the first roll,probability that the next roll will be a 5/96 ANSWER7 [assuming the die is again picked at random from the drawer]

DONE