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CUET · MATHS · PYQ PAPER 2025

If the random variable X has the following probability distribution :
X012Otherwise
P(X)K3K5K0
then \(K\) is equal to

  1. A \(\frac{4}{9}\)
  2. B \(\frac{2}{9}\)
  3. C \(\frac{1}{9}\)
  4. D \(\frac{8}{9}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{1}{9}\)

Step-by-step Solution

Detailed explanation

K + 3K + 5K = 1 9K = 1 K = \(\frac{1}{9}\)
From CUET
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