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Question

A coin is tossed. If head appears a fair die is thrown three times otherwise a biased die with probability of obtaining an even number twice as that of an odd number is thrown three times. If (n1,n2,n3) is an outcome, (1n1,n2,n36) and is found to satisfy the equation in1+in2+in3=1, then the probability that the fair die was thrown is (where i=1)

A
112
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B
13
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C
2759
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D
None of these
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Solution

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)
=1P({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

B12(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

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