MCQ
A random variable X has the following 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

Find the events E = {X : X is a prime number}, F{X : X < 4}, the probability $\text{P}(\text{E}\cup\text{F})$ is:

  • A
    0.50
  • B
    0.77
  • C
    0.35
  • D
    0.87

Answer

  1. 0.77

Solution:

P(E) = P(2) + P(3) + P(5) + P(7)

P(E) = 0.23 + 0.12 + 0.20 + 0.07

P(E) = 0.62

And

P(F) = P(1) + P(2) + P(3)

P(F) = 0.15 + 0.23 + 0.12

P(F) = 0.5

Also,

$\text{P}(\text{E}\cap\text{F})=\text{P}(2)+\text{P}(3)$

$\text{P}(\text{E}\cap\text{F})=0.23+0.12$

$\text{P}(\text{E}\cap\text{F})=0.35$

$\text{P}(\text{E}\cup\text{F})=\text{P}(\text{E})+\text{P(F)}-\text{P}(\text{E}\cap\text{F})$

$\text{P}(\text{E}\cup\text{F})=0.62+0.5-0.35$

$\text{P}(\text{E}\cup\text{F})=0.77$

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