An unbalanced dice (with 6 faces, numbered from 1 to 6) is thrown. The probability that the face value is odd is 1/2 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 B, which one of the following options is correct? A. 5/19 B. 2/3 C. 19/30 D. 1/6

GATE 2009 · Engineering Mathematics · Conditional Probability · medium

Answer: P(even | face > 3) = 2/3. Answer: B.

1. Find P(even) and P(odd): (1/2)*P(even) + P(even) = 1 => (3/2)*P(even) = 1 => P(even) = 2/3. So P(odd) = 1/3.
2. Find probability of each even face: P(2) = P(4) = P(6) = (2/3)/3 = 2/9.
3. Compute P(face > 3): P(4) = 2/9, P(6) = 2/9. For P(5): the three odd faces sum to 1/3 but their individual probabilities are not given as equal. However, we only need P(face > 3) to evaluate P(even and face > 3)/P(face > 3). P(even and face > 3) = P(4)+P(6) = 2/9+2/9 = 4/9.
4. Apply conditional probability: P(even and face > 3) = P(4)+P(6) = 4/9. P(face > 3) = P(4)+P(5)+P(6) = 2/9+2/9+2/9 = 6/9 = 2/3. So P(even | face > 3) = (4/9) / (2/3) = (4/9)*(3/2) = 12/18 = 2/3.