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 .…

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 PE∪F is
  1. 0.50
  2. 0.77
  3. 0.35
  4. 0.87

Solution

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

Asked in: AP EAMCET 2021 (19 Aug Shift 1)

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