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CUET · MATHS · PYQ PAPER 2023

A random variable X has the following distribution:
X01234567
P(X)03kk2k\(3k^2\)\(k^2\)\(6k^2\)3k
The value of k is:

  1. A \(\frac{1}{10}\)
  2. B -1
  3. C \(-\frac{1}{10}\)
  4. D \(\frac{1}{9}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{1}{10}\)

Step-by-step Solution

Detailed explanation

\( \sum P(X) = 1 \) \( 0 + 3k + k + 2k + 3k^2 + k^2 + 6k^2 + 3k = 1 \) \( 10k^2 + 9k - 1 = 0 \) \( (10k - 1)(k + 1) = 0 \) \( k = \frac{1}{10} \text{ or } k = -1 \) Since \( P(X) \ge 0 \), \( k = \frac{1}{10} \).
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