A random variable X has the probability distribution For the events E={xisprimenumber} and F={x<4} the probability of P(E∪F) is
A
0.50
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B
0.77
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C
0.35
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D
0.87
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Solution
The correct option is B0.77 Given, E={xisaprimenumber} 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