Questions: Probability Mass Functions and Discrete Distributions

Question 1

A discrete random variable X has PMF: p(1) = 0.2, p(2) = 0.3, p(3) = 0.4, p(4) = 0.1. What is P(X ≥ 3)?

• A. 0.4
• B. 0.5
• C. 0.6
• D. 0.9

Answer: (B)

P(X ≥ 3) = p(3) + p(4) = 0.4 + 0.1 = 0.5. This is computed by summing the PMF values at all outcomes in the event — the defining operation for discrete distributions. Option A (0.4) is p(3) alone, a classic off-by-one error from misreading 'at least 3' as 'exactly 3.'

Question 2

For a continuous random variable Y and a discrete random variable X, both describing the same phenomenon, which statement is always true?

• A. P(X = 5) can be positive; P(Y = 5) must equal zero
• B. P(X = 5) and P(Y = 5) are both positive if 5 is the most likely value
• C. P(X = 5) = 0 because individual points have no width
• D. Both P(X = 5) and P(Y = 5) must equal zero for the distributions to integrate to 1

Answer: (A)

The key distinction between discrete and continuous distributions: a discrete random variable can assign positive probability to individual values — P(X = 5) > 0 is perfectly valid and is exactly what the PMF specifies. For a continuous random variable, every single point has probability zero; probability only accumulates over intervals. Option B is a common misconception — the most likely value in a continuous distribution still has probability zero.

Question 3

A PMF can validly assign probability 0 to some values in the random variable's range, as long as all probabilities are nonnegative and sum to 1.

• T. True
• F. False

Answer: True

True. The two validity conditions for a PMF are p(x) ≥ 0 for all x and ∑p(x) = 1. Assigning p(x) = 0 to some values satisfies both conditions — a zero probability just means that outcome never occurs. For example, a die that never lands on 6 would have p(6) = 0 while still being a valid PMF.

Question 4

For a discrete random variable, most PMF values is expected to be equal — each outcome gets the same share of the probability budget.

• T. True
• F. False

Answer: False

False. Equal probabilities (a uniform distribution) are a special case, not a requirement. A valid PMF only requires nonnegativity and that all values sum to 1. A loaded die, for example, might have p(6) = 0.5 and p(1) = p(2) = p(3) = p(4) = p(5) = 0.1 — a completely valid PMF where outcomes have unequal probabilities.

Question 5

Why can a continuous random variable not have a probability mass function, and why does this distinction matter for computing event probabilities?

Model answer: A continuous random variable takes values on an uncountable range (e.g., all real numbers in an interval). If we tried to assign positive probability to each of uncountably many points, even tiny values would sum to infinity, violating the normalization condition. For continuous variables, probability accumulates over intervals, not points — P(a ≤ Y ≤ b) > 0 even when P(Y = a) = 0. For a discrete variable, event probabilities are computed by summing PMF values, which works because there are only countably many possible outcomes.

This distinction drives the entire difference in mathematical machinery: PMFs and sums for discrete distributions, probability density functions and integrals for continuous ones. Understanding that individual points carry zero probability in the continuous case prevents the common error of trying to compute P(Y = 5) for a normally distributed variable.