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CUET · MATHS · PYQ PAPER 2025

A random variable \(X\) has the following probability distribution:
X01234567
P(X)0k2k2k3k\(k^2\)\(2 k^2\)\(7 k^2+k\)
The value of \(P(4 < X < 7)\) is equal to

  1. A \(\frac{3}{100}\)
  2. B \(\frac{1}{2}\)
  3. C \(\frac{5}{100}\)
  4. D \(\frac{7}{100}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{3}{100}\)

Step-by-step Solution

Detailed explanation

\( \sum P(X=x) = 1 \) \( 0 + k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^2 + k) = 1 \) \( 10k^2 + 9k - 1 = 0 \) \( (10k-1)(k+1) = 0 \) \( k = \frac{1}{10} \) (since \(k\) must be non-negative) \( P(4 \( P(4…