    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 C 80171Let 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)=1019⇒P(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)0.75⇒P(face>3)=1057+10570.75=80171  Suggest Corrections  0      Similar questions  Explore more