An unbalanced dice (with 6 faces, numbered from 1 to 6) is thrown. The…
2009
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
Attempted by 4 students.
Show answer & explanation
Correct answer: B
Let P(even) = E. It is given that P(odd) = 0.9·E, so
E + 0.9E = 1 ⇒ 1.9E = 1 ⇒ E = 1/1.9 = 10/19 ≈ 0.5263158.
The three even faces (2, 4, 6) have equal probability, so each has probability
P(4) = P(6) = P(2) = E/3 = (10/19)/3 = 10/57 ≈ 0.1754386.
We are given P(even | >3) = 0.75. The outcomes greater than 3 are {4, 5, 6}. The even outcomes among these are 4 and 6, so
P(even ∩ >3) = P(4) + P(6) = 2 · (10/57) = 20/57.
Use the conditional probability formula:
0.75 = P(even | >3) = P(even ∩ >3) / P(>3) = (20/57) / P(>3).
Solve for P(>3): P(>3) = (20/57) / 0.75 = (20/57) · (4/3) = 80/171 ≈ 0.4678363.
Therefore the probability that the face value exceeds 3 is 80/171 ≈ 0.4678, which is closest to 0.468.