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Question

An unbalanced dice (with six faces numbered from 1 to 6) is thrown. The probability that the face value is odd is 90 % of the probability that the face value is even. The probability of getting any even numbered face is the same. If the probability that the face is even, given that it is greater than 3 is 0.75, then the probability that the face value exceeds 3, is

A
82171
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B
2957
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C
919
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D
80171
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Solution

The correct option is D 80171
Let E: face is even
and F: face is odd.
Let P(E)=p and P(F)=0.9p

P(E)+P(F)=1
P(E)=1019P(2)=P(4)=P(6)=1057

It is given that
P(E | face>3)=0.75
P(E face>3)P(face>3)=0.75
P(face=4,6)P(face>3)=0.75
P(face>3)=P(face=4,6)0.75
P(face>3)=P(4)+P(6)0.75P(face>3)=1057+10570.75=80171

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