GATE CS 2009 Q21 — Conditional Probability and Bayes Theorem

Let the six faces of the unbalanced die be represented by the set $S = \{1, 2, 3, 4, 5, 6\}$.

Let $O$ be the event of getting an odd face, $O = \{1, 3, 5\}$, and $E$ be the event of getting an even face, $E = \{2, 4, 6\}$.

We are given that the probability of getting an odd face is $90\%$ of getting an even face:

Since $P(O) + P(E) = 1$, we have:

Since the probability of getting any even numbered face is the same:

Let $G$ be the event that the face value is greater than $3$, i.e., $G = \{4, 5, 6\}$.

The intersection of even faces and faces greater than $3$ is:

Thus, the probability of getting an even face greater than $3$ is:

We are given the conditional probability that the face is even given that it is greater than $3$ is $0.75$:

Using the definition of conditional probability:

The value closest to $0.4678$ among the given options is $0.468$.