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Q.

A random variable X has the probability distribution 

X12345678
P(X)0.150.230.120.100.200.080.070.05

For the events E = {X:X is prime number} and F = {X:X < 4}, then the probability  P(E∪F)  is

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a

0.35

b

0.77

c

0.87

d

0.50

answer is B.

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Detailed Solution

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Required probability
=P(E∪F)=P({1,2,3}∪{2,3,5,7})=P({1,2,3,5,7})
=P(1)+P(2)+P(3)+P(5)+P(7) =0.15+0.23+0.12+0.20+0.07=0.77
 

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