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An unbalanced dice (with 6 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, which one of the following options is closest to the probability that the face value exceeds 3?

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Ans, 0.468

Let P_{n }be the probability of corresponding face value occurence.

Total probability

P_{1} + P_{2} + P3 + P_{4} + P_{5} + P_{6} = 1 .......... Eq. 1

According to conditions,

P_{1} + P3 + P_{5 }= 0.9( P_{2 }+ P_{4} + P_{6 }) ............... Eq. 2

P_{4} + P_{6 }= 0.75( P_{4} + P_{5} + P_{6} ) ..................... Eq. 3

but P_{2 }= P_{4} = P_{6 }= P given

So Eq, 1 P_{1} + P3 + P_{5 }= 1 - 3P_{2}

Eq. 2 P_{1} + P3 + P_{5 }= 0.9 * 3P_{2}

Soving these, P_{2 }= P_{4} = P_{6 }= P = 0.1754

Putting it in Eq. 3 P_{4} + P_{5} + P_{6} = 0.468

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