Question

Let \( X \) be a discrete random variable with \( P(X \in \{-5, -3, 0, 3, 5\}) = 1 \). Suppose that \( P(X = -3) = P(X = -5) \), \( P(X = 3) = P(X = 5) \) and \( P(X > 0) = P(X = 0) = P(X < 0) \). Then the value of \( 12P(X = 3) \) is equal to __________ (answer in integer).

Answer 1

Correct Answer: 2

1. Define Probabilities: - Let \( P(X = -5) = P(X = -3) = p \). - Let \( P(X = 3) = P(X = 5) = q \). - Let \( P(X = 0) = r \). 2. Given Conditions: - From \( P(X > 0) = P(X = 0) = P(X < 0) \): \[ q + q = r = p + p. \] - Simplify: \[ 2q = r = 2p \implies q = p. \] 3. Total Probability: - Since \( P(X \in \{-5, -3, 0, 3, 5\}) = 1 \): \[ 2p + r + 2q = 1. \] - Substituting \( r = 2p \) and \( q = p \): \[ 2p + 2p + 2p = 1 \implies 6p = 1 \implies p = \frac{1}{6}. \] - Thus: \[ q = \frac{1}{6}. \] 4. Value of \( 12P(X = 3) \): - \( P(X = 3) = q = \frac{1}{6} \). 
Multiply by 12: \[ 12P(X = 3) = 12 \cdot \frac{1}{6} = 2. \]