Question

there are 3 dice.

the first dice is fair

the second dice has probability P(1)=P(2)=P(3)=1/9

P(4)=P(5)=P(6)=2/9

the third has probability

p(1)=p(2)=p(3)=p(4)=1/9

p(5)=1/3

p(6)=2/9

if you select one random dice and roll it, and A is the event the fair dice was picked and B the event the result is 6. knowing you rolled a 6, what is the outcome you chose the fair dice?

Answer 1

Event B = rolling a 6

Event A = rolling die 1

Event 2 = rolling die 2

Event 3 = rolling die 3

P[ B | A ] = 1/6 ( fair die )

P[ B | 2 ] = 2/9

P[ B | 3 ] = 2/9

P[ A ] = P[ 1 ] =P[ 2 ] = 1/3

P[ B ] = P[ B | A ]*P[ A ] + P[ B | 2 ]*P[ 2 ] + P[ B | 3 ]*P[ 3 ]

P[ B ] = ( 1/6)*(1/3) + (2/9)*(1/3) + (2/9)*(1/3)

P[ B ] = 1/18 + 2/27 + 2/27

P[ B ] = 1/18 + 4/27

P[ B ] = 11/54

We need to find

P[ A | B ] = P[ B | A ]*P[ A ] / P[ B ]

P[ A | B ] = ( 1/6)*(1/3) / 11/54

P[ A | B ] = (1/18)/(11/54)

P[ A | B ] = 3/11