A random variable X has the probability distribution
For the events E = {X is a prime number} abd `F={Xlt4},P(EcupF)` is

Options

A. 0.77
B. 0.87
C. 0.35
D. 0.5

Answer

A

Explanation

`P(E)=P(X=2)+P(X=3)+P(X=5)+P(X=7)`
`=0.23+0.12+0.20+0.07=0.62`
`P(F)=P(X=1)+P(X=2)+P(X=3)`
`=0.15+0.23+0.12=0.5`
`P(EcapF)=P(X=2)+P(X=3)=0.23+0.12=0.35`
`therefore" "P(EcupF)=P(E)+P(F)-P(EcapF)`
`=0.62+0.5-0.35=0.77`