Question

Let a random variable X take values 0, 1, 2, 3 with $ P(X = 0) = P(X = 1) = p, \, P(X = 2) = P(X = 3), \, \text{and} \, F(X^2) = 2F(X). $ Then the value of $ 8p - 1 $ is:

• A. 0
• B. 2
• C. 1
• D. 3

Hint:

For random variable problems, carefully use the total probability condition and the cumulative distribution functions (CDF) for solving.

Answer 1

The Correct Option is B

Let \( P(X = 2) = P(X = 3) = q \). Thus, the total probability condition gives: \[ P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 1 \] \[ p + p + q + q = 1 \] \[ 2p + 2q = 1 \quad \Rightarrow \quad p + q = \frac{1}{2} \quad \cdots (1) \] Next, we know that \( F(X^2) = 2F(X) \). The cumulative distribution function (CDF) is defined as: \[ F(X) = P(X \leq x) \] From the problem, we have: \[ F(1) = P(X \leq 1) = P(X = 0) + P(X = 1) = p + p = 2p \] \[ F(4) = P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = p + p + q + q = 2p + 2q \] From the condition \( F(X^2) = 2F(X) \), we substitute: \[ F(4) = 2F(1) \] Thus, \[ 2p + 2q = 2(2p) \] \[ 2p + 2q = 4p \] \[ 2q = 2p \] \[ q = p \] Substituting \( q = p \) into equation (1): \[ p + p = \frac{1}{2} \] \[ 2p = \frac{1}{2} \] \[ p = \frac{1}{4} \] Finally, we compute \( 8p - 1 \): \[ 8p - 1 = 8 \times \frac{1}{4} - 1 = 2 - 1 = 1 \] Thus, the value of \( 8p - 1 \) is \( 2 \).