Mathematics - Probability Question with Solution | TestHub

MathematicsProbabilityRandom Variable and its Probability DistributionEasy2 minPYQ_2025

MathematicsEasysingle choice

Let a random variable X take values $0, 1, 2, 3$ with $P (X = 0) = P (X = 1) = p, P (X = 2) = P (X = 3)$ and $E (X^{2}) = 2 E (X)$ . Then the value of $8 p - 1$ is :

Options:

Answer:

Solution:

$2 p + 2 q = \frac{1}{2} p + q E (x^{2}) = i = 0 \sum 3 x_{i}^{2} p (x_{i}) = 0 \cdot p + 1 \cdot p + 4 \cdot q + 9 q = p + 13 q E (x) = i = 0 \sum 3 x_{i}^{2} p (x_{i}) = 0 \cdot p + 1 \cdot p + 2 q + 3 q = p + 5 q p + 13 q = 2 (p + 5 q) p = 3 q So, q = \frac{1}{8} & p = \frac{3}{8} So, 8 p - 1 = 2 Option (2)$

Stream:JEESubject:MathematicsTopic:ProbabilitySubtopic:Random Variable and its Probability Distribution

⏱ 2mℹ️ Source: PYQ_2025

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