Word Count : 668

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We are given a bivariate probability distribution \( P(X = x, Y = y) = \frac{x^2 + y}{32} \) where \( x = 0, 1, 2, 3 \) and \( y = 0, 1 \). Let’s solve for the four parts of the question step by step: ### (i) \( P(X \leq 1, Y = 1) \) This is the joint probability that \( X \leq 1 \) and \( Y = 1 \), meaning we need to sum the probabilities for \( X = 0, 1 \) and \( Y = 1 \). \[ P(X \leq 1, Y = 1) = P(X = 0, Y = 1) + P(X = 1, Y = 1) \] Substituting the formula for the joint probability: \[ P(X = 0, Y = 1) = \frac{0^2 + _______ _____ ___ _________ _________ _________ ___ ______ _______ _____ ____ ____.
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