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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%. 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?
0.6
0.468
0.5
1
given p(odd)=0.9 p(even)
∑ p(x)=1
P(odd) +p(even)=1
0.9 p(even)+p(even)=1
P(even )=1/1.9
=0.5263
It is given that p is same
P(2)=p(4)=p(6)
P (even)=p(2) or p(4) or p(6 )=p(2)+p(4)+p(6)
P(2)=p(4)=p(6)=1/3 p(even)
=1/3 (0.5263)=0.1754
Given that p(even |face>3)=0.75
P(even ?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(4) +p(6)/0.75
=(0.1754 +0.1754)/0.75
=0.468
By: bhavesh kumar singh ProfileResourcesReport error
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