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AP EAMCET · Maths · Probability

A random variable X has the probability distribution :
X 1 2 3 4 5 6 7 8
PX 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 PEF is

  1. A 0.50
  2. B 0.77
  3. C 0.35
  4. D 0.87
Verified Solution

Answer & Solution

Correct Answer

(B) 0.77

Step-by-step Solution

Detailed explanation

PE=P2 or 3 or 5 or 7 =0.23+0.12+0.20+0.07=0.62 PF=P1 or 2 or 3 =0.15+0.23+0.12=0.50 PE∩F=P2 or 3 =0.23+0.12=0.35 ∴PEUF=PE+PF−PE∩F =0.62+0.50−0.35=0.77