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CUET · MATHS · PYQ PAPER 2025

The probability distribution of a random variable X is given by
X012
P(X)\(1-7 a^2\)\(\frac{1}{2} a+\frac{1}{4}\)\(a^2\)
If \(a>0\), then \(P(0<X \leq 2)\) is equal to

  1. A \(\frac{1}{16}\)
  2. B \(\frac{3}{18}\)
  3. C \(\frac{7}{16}\)
  4. D \(\frac{9}{16}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{7}{16}\)

Step-by-step Solution

Detailed explanation

\( (1-7a^2) + (\frac{1}{2}a+\frac{1}{4}) + a^2 = 1 \) \( -6a^2 + \frac{1}{2}a + \frac{1}{4} = 0 \) \( 24a^2 - 2a - 1 = 0 \) \( a = \frac{2 + \sqrt{(-2)^2 - 4(24)(-1)}}{2(24)} = \frac{2 + \sqrt{4+96}}{48} = \frac{2 + 10}{48} = \frac{1}{4} \) \( P(0 \( P(0 \( P(0…
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