Random Variable And Expectation
A discrete random variable X takes values 0, 1, and 2 with respective probabilities 0.2, 0.5, and 0.3 in an experiment; compute the expected value E(X) of this random variable.
Select the correct option:
Solution
1.1
The expectation of a discrete random variable is the probability-weighted average of its values, defined as E(X) = Σ xᵢ P(xᵢ), representing the long-run mean outcome. This expectation computation is a fundamental JEE Advanced random-variable skill. The probabilities 0.2, 0.5, 0.3 sum to 1, confirming a valid distribution. Then E(X) = 0 × 0.2 + 1 × 0.5 + 2 × 0.3 = 0 + 0.5 + 0.6 = 1.1. Option 1.0 results from treating the values as equally likely and averaging 0, 1, 2. Option 1.3 comes from misassigning probability 0.5 to the value 2. Option 0.9 arises from a sign or weighting slip in summing the products. The principle is the definition of expectation as a weighted mean, where each value is weighted by its probability of occurrence rather than being treated uniformly. This distinction matters because expectation need not equal any actually attainable value of the variable; it is a balance point of the distribution, much like a centre of mass for the probability weights placed at each value. Plausibility check: E(X) = 1.1 lies inside the value range [0, 2] and sits near the most probable value 1, which carries the largest probability 0.5, so the answer is consistent with where the distribution's mass concentrates, and it slightly exceeds 1 because the value 2 still carries appreciable weight 0.3.
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About This Question
- Subject
- mathematics
- Chapter
- statistics and probability
- Topic
- random variable and expectation
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
1.1
The expectation of a discrete random variable is the probability-weighted average of its values, defined as E(X) = Σ xᵢ P(xᵢ), representing the long-run mean outcome. This expectation computation is a fundamental JEE Advanced random-variable skill. The probabilities 0.2, 0.5, 0.3 sum to 1, confirming a valid distribution. Then E(X) = 0 × 0.2 + 1 × 0.5 + 2 × 0.3 = 0 + 0.5 + 0.6 = 1.1. Option 1.0 results from treating the values as equally likely and averaging 0, 1, 2. Option 1.3 comes from misassigning probability 0.5 to the value 2. Option 0.9 arises from a sign or weighting slip in summing the products. The principle is the definition of expectation as a weighted mean, where each value is weighted by its probability of occurrence rather than being treated uniformly. This distinction matters because expectation need not equal any actually attainable value of the variable; it is a balance point of the distribution, much like a centre of mass for the probability weights placed at each value. Plausibility check: E(X) = 1.1 lies inside the value range [0, 2] and sits near the most probable value 1, which carries the largest probability 0.5, so the answer is consistent with where the distribution's mass concentrates, and it slightly exceeds 1 because the value 2 still carries appreciable weight 0.3.
This medium difficulty mathematics question is from the chapter statistics and probability, covering the topic of random variable and expectation. It appeared in the 2025 exam.
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