A random variable $X$ has the following probability distribution:

$X:$	1	2	3	4	5	6	7	8
$p\left(X\right):$	0.15	0.23	0.12	0.10	0.20	0.08	0.07	0.05

For the events $E=\left\{X\mathrm{is}\mathrm{a}\mathrm{prime}\mathrm{number}\right\}$ and $F=\{X<4\},$ then the value of $100P(E\cup F)$ is

$E=\left\{xisaprimenumber\right\}=\{2,3,5,7\}$

$P\left(E\right)=P(X=2)+P(X=3)+P(X=5)+P(X=7)=0.23+0.12+0.20+0.07=0.62$

$F=\{X<4\}=\{1,2,3\}$

$P\left(F\right)=P(X=1)+P(X=2)+P(X=3)=0.15+0.23+0.12=0.5$

$E\cap F=\left\{Xisprimeaswellaslessthan4\right\}=\{2,3\}$

$P(E\cap F)=P(X=2)+P(X=3)=0.23+0.12=0.35$

Therefore, the required probability is 

$P(E\cup F)=P\left(E\right)+P\left(F\right)-P(E\cap F)=0.62+0.5-0.35=0.77$