Key Concepts and Formulas

1. Probability Mass Function (PMF) of a Discrete Random Variable: For a discrete random variable $X$ taking values $x_i$, its PMF is defined by $P(X=x_i)$. The sum of all probabilities must equal 1 (Normalization Property): $\sum P(X=x_i) = 1$. Also, each individual probability must be between 0 and 1, inclusive ($0 \le P(X=x_i) \le 1$).

2. Expected Value (Mean) of a Discrete Random Variable $X$: The expectation, denoted $E(X)$, is the weighted average of all possible values of $X$, where the weights are their respective probabilities. $E(X) = \sum x_i P(X=x_i)$

3. Expected Value of a Function of a Discrete Random Variable $g(X)$: If $g(X)$ is any function of $X$ (e.g., $X^2$), its expected value is calculated by applying the function to each value $x_i$ and multiplying by its probability, then summing these products. $E(g(X)) = \sum g(x_i) P(X=x_i)$ For $E(X^2)$, we use $g(X)=X^2$: $E(X^2) = \sum x_i^2 P(X=x_i)$

---

Step-by-Step Solution

Step 1: Define the Probability Mass Function (PMF) for $X$ We begin by clearly listing all possible values of the random variable $X$ and their corresponding probabilities based on the information given in the problem. This initial setup is crucial for constructing the equations.

The random variable $X$ takes values $0, 1, 2, 3$.

• We are given $P(X=0) = p$.
• We are given $P(X=1) = p$.
• We are given $P(X=2) = P(X=3)$. Let's introduce a new variable, say $q$, for these probabilities. So, $P(X=2) = q$.
• Consequently, $P(X=3) = q$.

Step 2: Apply the Normalization Property of PMF to Form the First Equation A fundamental property of any probability distribution is that the sum of probabilities for all possible outcomes must be equal to 1. This property allows us to establish the first algebraic relationship between $p$ and $q$.

Summing all probabilities: $P(X=0) + P(X=1) + P(X=2) + P(X=3) = 1$ Substitute the probabilities defined in Step 1: $p + p + q + q = 1$ Combine like terms: $2p + 2q = 1 \quad \text{(Equation 1)}$

Step 3: Calculate the Expected Value of $X$, $E(X)$ The problem provides a condition involving $E(X)$, so we must calculate its expression in terms of $p$ and $q$. We use the formula $E(X) = \sum x_i P(X=x_i)$.

$E(X) = (0 \cdot P(X=0)) + (1 \cdot P(X=1)) + (2 \cdot P(X=2)) + (3 \cdot P(X=3))$ Substitute the probabilities: $E(X) = (0 \cdot p) + (1 \cdot p) + (2 \cdot q) + (3 \cdot q)$ Simplify the expression: $E(X) = 0 + p + 2q + 3q$ $E(X) = p + 5q \quad \text{(Equation 2)}$

Step 4: Calculate the Expected Value of $X^2$, $E(X^2)$ The given condition also involves $E(X^2)$, so we need to calculate its expression. We use the formula $E(X^2) = \sum x_i^2 P(X=x_i)$. It is important to square the values of $X$ ($x_i$) before multiplying by their probabilities.

$E(X^2) = (0^2 \cdot P(X=0)) + (1^2 \cdot P(X=1)) + (2^2 \cdot P(X=2)) + (3^2 \cdot P(X=3))$ Substitute the probabilities: $E(X^2) = (0 \cdot p) + (1 \cdot p) + (4 \cdot q) + (9 \cdot q)$ Simplify the expression: $E(X^2) = 0 + p + 4q + 9q$ $E(X^2) = p + 13q \quad \text{(Equation 3)}$

Step 5: Use the Given Condition to Form the Second Equation The problem statement provides a crucial relationship: $E(X^2) = 2E(X)$. We can substitute the expressions for $E(X)$ and $E(X^2)$ that we derived in the previous steps into this condition to form our second independent equation relating $p$ and $q$.

Substitute Equation 2 and Equation 3 into the given condition: $(p + 13q) = 2(p + 5q)$ Distribute the 2 on the right side: $p + 13q = 2p + 10q$ Rearrange the terms to isolate $p$ in terms of $q$: $13q - 10q = 2p - p$ $3q = p \quad \text{or} \quad p = 3q \quad \text{(Equation 4)}$

**Step 6: Solve the System of Linear Equations for $p$ and $q$} Now we have a system of two linear equations with two variables:

1. $2p + 2q = 1$ (from Step 2)
2. $p = 3q$ (from Step 5)

We can use the substitution method. Substitute Equation 4 ($p=3q$) into Equation 1: $2(3q) + 2q = 1$ $6q + 2q = 1$ $8q = 1$ $q = \frac{1}{8}$ Now substitute the value of $q$ back into Equation 4 to find $p$: $p = 3 \left(\frac{1}{8}\right)$ $p = \frac{3}{8}$ We have successfully determined the values of $p$ and $q$.

Step 7: Calculate the Value of $8p-1$ The final step is to calculate the value of the expression requested in the problem, $8p-1$, using the value of $p$ we just found.

Substitute $p = 3/8$ into the expression: $8p - 1 = 8 \left(\frac{3}{8}\right) - 1$ $8p - 1 = 3 - 1$ $8p - 1 = 2$

---

Common Mistakes & Tips

• Confusing $E(X^2)$ and $(E(X))^2$: Always remember that $E(X^2) = \sum x_i^2 P(X=x_i)$ while $(E(X))^2 = (\sum x_i P(X=x_i))^2$. These are generally not equal.
• Forgetting Normalization: The sum of all probabilities must always be 1. This is a crucial equation in many probability problems.
• Verification: After finding $p$ and $q$, quickly check if all probabilities are between 0 and 1, and if their sum is 1. This helps catch algebraic errors early. For example, $p=3/8$, $q=1/8$. All are valid probabilities, and $2(3/8) + 2(1/8) = 6/8 + 2/8 = 8/8 = 1$.

---

Summary

This problem required a systematic application of the fundamental concepts of discrete probability distributions. We began by defining the probability mass function using the given information and introducing an auxiliary variable $q$. We then used the normalization property of probabilities to form the first equation. Next, we calculated the expected values $E(X)$ and $E(X^2)$ using their respective definitions. The given condition $E(X^2) = 2E(X)$ provided the second equation, allowing us to solve the system of linear equations for $p$ and $q$. Finally, we substituted the calculated value of $p$ into the target expression $8p-1$ to arrive at the solution.

The final answer is $\boxed{2}$, which corresponds to option (A).