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AP EAMCET202119 Aug 2021Morning ShiftMathematicsProbabilityActual

A random variable X has the probability distribution : X 1 2 3 4 5 6 7 8 P X 0 . 15 0 . 23 0 . 12 0 . 10 0 . 20 0 . 08 0 . 07 0 . 05 For the events E = X is a prime number and F = X < 4 , then P E ∪ F is

Options

  1. A0 . 50
  2. B0 . 77
  3. C0 . 35
  4. D0 . 87

Correct answer

B. 0 . 77

Step-by-step solution

P E = P 2   or   3   or   5   or   7 = 0.23 + 0.12 + 0.20 + 0.07 = 0.62 P F = P 1   or   2   or   3 = 0.15 + 0.23 + 0.12 = 0.50 P E ∩ F = P 2   o r   3 = 0.23 + 0.12 = 0.35 ∴ P E U F = P E + P F − P E ∩ F = 0.62 + 0.50 − 0.35 = 0.77

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