PMF Joint and Marginal PMFs 3

Question 1

Let X and Y have joint PMF p(x, y) = 1/6 for (x, y) ∈ {(1,1),(1,2),(2,1),(2,2),(3,1),(3,2)}. What is p_Y(1)?

Accepted Answer

1/2

Answer

p_Y(1) = p(1,1) + p(2,1) + p(3,1) = 1/6 + 1/6 + 1/6 = 3/6 = 1/2.

Question 2

Which statement correctly describes the support of a joint PMF?

Accepted Answer

The set of pairs (x, y) where p(x, y) > 0

Answer

The support of a joint PMF is the set of all pairs (x, y) for which p(x, y) > 0, i.e., the pairs that have positive probability.

Question 3

For independent X ~ Binomial(2, 0.5) and Y ~ Bernoulli(0.5), what is p(X=1, Y=1)?

Accepted Answer

0.25

Answer

By independence, p(X=1,Y=1) = P(X=1)·P(Y=1) = C(2,1)(0.5)²·0.5 = 0.5·0.5 = 0.25.

Question 4

Which of the following joint PMF tables corresponds to X and Y being independent?

Accepted Answer

p(1,1)=0.06, p(1,2)=0.14, p(2,1)=0.24, p(2,2)=0.56

Answer

In option B, marginals are p_X(1)=0.2, p_X(2)=0.8, p_Y(1)=0.3, p_Y(2)=0.7, and each cell equals p_X(x)·p_Y(y), confirming independence.

Question 5

If X and Y are discrete random variables, E[g(X, Y)] is computed as:

Accepted Answer

∑_x ∑_y g(x, y) · p(x, y)

Answer

The expected value of any function g(X, Y) is ∑_x ∑_y g(x, y) · p(x, y), weighting each value by its joint probability.

PMF - Probability Mass Function Practice Test