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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?
asked in Probability by gate

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Ans, 0.468

Let Pbe the probability of corresponding face value occurence.

Total probability

P1 + P2 + P3 + P4 + P5 + P6 = 1 .......... Eq. 1

According to conditions,

P1 + P3 + P= 0.9( P+ P4 + P) ............... Eq. 2

P4 + P= 0.75( P4 + P5 + P6 ) ..................... Eq. 3

but P2 = P4 = P= P given

So Eq, 1  P1 + P3 + P= 1 - 3P2

Eq. 2  P1 + P3 + P= 0.9 * 3P2

Soving these,  P= P4 = P= P = 0.1754

Putting it in Eq. 3  P4 + P5 + P6 = 0.468

answered by Madhav Goratela

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