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MHT CET2008MathematicsProbability

A random variable X has the probability distribution array l|c|c|c|c|c|c|c|c & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 (x) & 0.15 & 0.23 & 0.12 & 0.10 & 0.20 & 0.08 & 0.07 & 0.05 array For the events E= x is prime number and F= x < 4 the probability of 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

Given, E= x is a prime number aligned P(E) &=P(2)+P(3)+P(5)+P(7) &=0.23+0.12+0.20+0.07=0.62 and F &= x < 4 P(F) &=P(1)+P(2)+P(3) &=0.15+0.23+0.12=0.50 and P(E F) &=P(2)+P(3) =& 0.23+0.12=0.35 P(E F) &=P(E)+P(F)-P(E F) &=0.62+0.50-0.35 &=0.77 aligned

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