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
Answer: (B)
Explanation: Let Xi be the probability that the face value is i. We can say that,
X1 + X2 + X3 + X4 + X5 + X6 = 1
It is given that,
(X1 + X3 + X5) = 0.9*(X2 + X4 + X6)
It is also given that,
X2 = X4 = X6
Also given that,
(X4 + X6)/(X4 + X5 + X6) = 0.75
Solving the above equations, we can get,
X4 + X5 + X6 = 0.468
So, option (B) is correct.
Improvement done by maninuthi.
Quiz of this Question