Lets breakdown the question
=>An unbalanced dice (with 6 faces, numbered from 1 to 6) is thrown
Unbalanced dice means which the probability of getting a face is not equally likely when thrown( You can say each probability of each face cannot be equal)
And Let’s say P1, P2, P3, P4, P5 and P6 are probability of getting face 1 to 6 respectively.
=>The probability that the face value is odd is 90% of the probability that the face value is even.
It means P({1,3,5}) = 0.9 * P({2,4,6})
Let’ say P(Odd) = P({1,3,5}) , P({2,4,6}) = P(Even)
=>The probability of getting any even numbered face is the same.
P2 = P4 = P6
=>If the probability that the face is even given that it is greater than 3 is 0.75
P(Even / >3) = 0.75
=>which one of the following options is closest to the probability that the face value exceeds 3?
P(>3) = ? ( Have to find this)
Given Info:
Unbalanced dice
P(Odd) = P({1,3,5}) , P({2,4,6}) = P(Even) and P({1,3,5}) = 0.9 * P({2,4,6})
P2 = P4 = P6
P(Even / >3) = 0.75
P(>3) = ?
Solution :
P(Odd) + P (Even) = 1
0.9 P(Even) + P(Even) = 1
P(Even) = 1/1.9 ---->(1)
P(Even/ >3) = P(Even $\cap$ >3) / P(>3)
0.75 = P(Even $\cap$ >3) / P(>3)
We know that
P(A $\cap$ B) = P(A/B) P(A) = P(B/A) = P(B)
Therefore, P(>3) = P(Even) P( >3/Even) / 0.75
For P(>3/ Even) = Probability of >3 for a given even number (2,4,6) and they mentioned in the question that “The probability of getting any even numbered face is the same”. 2 out of 3 are >3. therefore it is 2/3.
P(>3) = [(1/1.9) (2/3)] / 0.75
P(>3) = 0.468 (Option B)