Q.A die is loaded in such a way that each odd number is twice as likely to occur as each even number. Find $P(G)$, where $G$ is the event that a number greater than 3 occurs on a single roll of the die.

The key idea here is assigning probabilities to individual outcomes in a non-uniform sample space.

1. Let $k$ be the probability of an even number occurring. According to the problem, the probability of an odd number occurring is $2k$.
2. The sample space for a single roll is $S = \{1, 2, 3, 4, 5, 6\}$. We assign probabilities to each outcome: $P(1) = 2k$, $P(2) = k$, $P(3) = 2k$, $P(4) = k$, $P(5) = 2k$, $P(6) = k$.
3. The sum of probabilities of all outcomes must be 1: $2k + k + 2k + k + 2k + k = 1 \implies 9k = 1 \implies k = \frac{1}{9}$. …

When dealing with a loaded die, the fundamental assumption of equally likely outcomes, which applies to a fair die, no longer holds. Instead, we are given a specific relationship between the probabilities of different outcomes. Our task is to use this relationship to determine the individual probabilities of each face and then use these to calculate the probability of a specific event.

The core idea is to:

1. Represent the unknown probabilities using a variable based on the given condition.
2. Use the axiom that the sum of probabilities of all possible outcomes in the sample space must equal 1 to solve for this variable. This step normalizes the relative probabilities into actual probabilities.
3. Once individual probabilities are known, calculate the probability of the desired event by summing the probabilities of the outcomes that constitute that event.

Let's break this down step-by-step.

1. Define the Sample Space

   When a standard die is rolled, the set of all possible outcomes, known as the sample space $S$, is:

   $S = \{1, 2, 3, 4, 5, 6\}$

2. Understand the Loading Condition

   The problem states that "each odd number is twice as likely to occur as each even number".

   The odd numbers in our sample space are $\{1, 3, 5\}$.

   The even numbers in our sample space are $\{2, 4, 6\}$.

   This means:

   $P(1) = P(3) = P(5)$

   $P(2) = P(4) = P(6)$

   And, crucially, for any specific odd number $o$ and any specific even number $e$:

   $P(o) = 2 \times P(e)$

3. Assign Relative Probabilities

   Let's assign a variable to the probability of an even number.

   Let $P(2) = P(4) = P(6) = x$.

   According to the loading condition, each odd number is twice as likely as each even number. So:

   $P(1) = P(3) = P(5) = 2x$.

4. Normalize Probabilities

   A fundamental axiom of probability states that the sum of the probabilities of all possible outcomes in the sample space must be equal to 1.

   So, we have:

   $P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1$

   Substitute the expressions in terms of $x$:

   $(2x) + (x) + (2x) + (x) + (2x) + (x) = 1$

   Combine the terms:

   $9x = 1$

   Solve for $x$:

   $x = \frac{1}{9}$

5. Calculate Individual Probabilities …