An unbalanced dice (with 6 faces, numbered from 1 to 6) is thrown. The...
Problem Statement
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?
Solution
Let's denote the probability of getting an even face as 'p'. Then, the probability of getting an odd face will be '9p/10', as given in the problem statement.
Calculating Probability of Getting an Even Face
The probability of getting any even numbered face is the same. Therefore, the probability of getting an even face will be the sum of probabilities of getting 2, 4, and 6, which are all even numbers. Hence,
p = (1/6) + (1/6) + (x/6) = (2/6) + (x/6) = (2+x)/6
where 'x' is the probability of getting any other even number apart from 2 and 4.
Calculating Probability of Getting an Even Face Given that it is Greater than 3
The probability of getting an even face given that it is greater than 3 is 0.75. Therefore,
P(even | greater than 3) = P(even and greater than 3) / P(greater than 3) = 0.75
=> P(even and greater than 3) = 0.75 * P(greater than 3)
=> P(even and 4 or 6) = 0.75 * (1/2) = 0.375
=> P(even and 4) + P(even and 6) = 0.375
=> (1/6) + x/6 = 0.375
=> x = 1.5
Substituting this value of 'x' in the equation for 'p', we get
p = (2+1.5)/6 = 0.5833
Calculating Probability of Getting a Face Exceeding 3
The probability of getting a face exceeding 3 is the sum of probabilities of getting 4, 5, and 6. Hence,
P(face exceeding 3) = P(4) + P(5) + P(6) = (1/6) + (1/6) + (1/6) = 0.5
Therefore, the closest option to the probability that the face value exceeds 3 is option 'B', i.e., 0.468.