The correct option is C 2759
Let E be an event F represents tossing of a fair die and B represents tossing of a biased die.
P(F|(n1,n2,n3)E)=P((n1,n2,n3)E|F)×P(F)P((n1,n2,n3)E|F)×P(F)+P((n1,n2,n3)E|B)×P(B)
P(F)=P(B)=12
P(1)=P(2)⋯=P(6)=16 (for a fair dice.)
2×P({1,3,5}=P({2,4,6}) (for biased dice)
=1−P({1,3,5})
P({1,3,5}=13
P({2,4,6})=23
Assume that 1,3,5 are equally likely for biased dice and same for 2,4,6
P(1)=P(3)=P(5)=19
P(2)=P(4)=P(6)=29in1+in2+in3=1
We can have 2 possible cases
(1,1,−1) or i,−i,1
Permutations of n1,n2,n3 will give
n1=4,n2=4,n3=(2,6)
n1,n2,n3→(4,4,2) or (4,4,6)
Each of these cases can be permutated in number of ways
=3!2!=3
For fair dice :
P(n1=4,n2=4,n3=(2,6)|F)=6×16×16×16=136
For biased die
(4,4,2) (4,4,6)
29×29×29 29×29×29
in1+in2+in3=14
(i, −i, 1)
n1 n2 n3
(1,5) 3 4
(1,3,4) (5,3,4)
6 ways 6 ways
Fair12(16×16×16)=236
B→12(19×19×29)=8243
P((n1,n2,n3)E/F)=336=112
P((n1,n2,n3)E/B)=336=16243+8243=24243
P(F/(n1,n2,n3)E)=112×12112×12+24243×12
=1427+3227×4=2759