Question:

Which of the following cannot be the probability mass function of any discrete random variable \(X\)?

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Always verify that the sum of probabilities across all values of the random variable equals 1 for a valid PMF.
Updated On: May 26, 2025
  • \(P(X = x) = \frac{1}{6} \text{ for } x = 1,2,3,4,5,6\)
  • \(P(X = x) = \frac{1}{3} \text{ for } x = 1,2,3\)
  • \(P(X = x) = \frac{1}{2} \text{ for } x = 1,2\)
  • \(P(X = x) = \frac{1}{4} \text{ for } x = 1,2\)
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The Correct Option is D

Solution and Explanation

A probability mass function (PMF) must satisfy the condition: \[ \sum P(X = x_i) = 1 \] Let us examine each option: Option A: \[ P(X = x) = \frac{1}{6} \text{ for } x = 1,2,3,4,5,6 \Rightarrow \sum P(X = x) = 6 \times \frac{1}{6} = 1 \quad \text{(Valid)} \] Option B: \[ P(X = x) = \frac{1}{3} \text{ for } x = 1,2,3 \Rightarrow \sum P(X = x) = 3 \times \frac{1}{3} = 1 \quad \text{(Valid)} \] Option C: \[ P(X = x) = \frac{1}{2} \text{ for } x = 1,2 \Rightarrow \sum P(X = x) = 2 \times \frac{1}{2} = 1 \quad \text{(Valid)} \] Option D: \[ P(X = x) = \frac{1}{4} \text{ for } x = 1,2 \Rightarrow \sum P(X = x) = 2 \times \frac{1}{4} = \frac{1}{2} \neq 1 \quad \text{(Invalid)} \] Hence, option (D) cannot be a valid PMF.
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