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himanshu2610 · Aug 9, 2009, 07:26 AM

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?

(A) 0.453 (B) 0.468 (C) 0.485 (D) 0.492

nareshguru · Jan 8, 2011, 09:39 AM

let probability that face value is even be x
probability that face value is even be 0.9x
x+0.9x=1 , x=1/1.9
p(face=2)=p(face=4)=p(face=6)=x/3
p(face value even|| face>3)=P(face value even && face>3)/P(face>3)=0.75
now, P(face value even && face>3)=p(face=4)+ p(face=6)=2x/3
then
(2x/3)/p(face>3)=0.75
P(face>3)=2x/3*0.75=2/3*0.75*1.9 {x=1/1.9}
P(face>3)=0.468

that's solve...