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CUET · MATHS · PYQ PAPER 2023

Probability distribution of random variable \(X\) is:
x\(0\)\(1\)\(2\)\(3\)\(4\)\(5\)
\(P(X)\)\[3k\]\(3k\)\(2k\)\(2k^2\)\(3k^2\)\(4k^2\)
The value of \(k\) is:

  1. A \(k=-1, \frac{1}{9}\)
  2. B \(k=-1\)
  3. C \(k=\frac{1}{9}\)
  4. D \(k=0,-9\)
Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

\(\sum P(X) = 1\) \(3k + 3k + 2k + 2k^2 + 3k^2 + 4k^2 = 1\) \(9k^2 + 8k - 1 = 0\) \((9k - 1)(k + 1) = 0\) \(k = \frac{1}{9}\) or \(k = -1\) For \(P(X) \ge 0\): If \(k = -1\), \(P(X=0) = 3(-1) = -3\), which is invalid. If \(k = \frac{1}{9}\), all \(P(X)\) values are non-negative.…
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