A random variable \(X\) has the following probability distribution

For the event \(E=\{X\) is a prime number \(\}\), \(F=\{X \lt 4\}\), then \(P(E \cup F)\) is

\(\begin{aligned} \mathrm{P}(\mathrm{E}) & =\mathrm{P}(\mathrm{X} \text { is a prime number }) \\ & =\mathrm{P}(\mathrm{X}=2 \text { or } \mathrm{X}=3 \text { or } \mathrm{X}=5 \text { or } \mathrm{X}=7) \\ & =\mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}=3)+\mathrm{P}(\mathrm{X}=5) \\ & +\mathrm{P}(\mathrm{X}=7) \\ & =0.23+0.10+0.20+0.07 \\ & =0.6\end{aligned}\)
\(\begin{aligned} \mathrm{P}(\mathrm{F}) & =\mathrm{P}(\mathrm{X} \lt 4) \\ & =\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}=3) \\ & =0.15+0.23+0.10 \\ & =0.48 \\ \mathrm{P}(\mathrm{E} \cap \mathrm{F}) & =\mathrm{P}(\mathrm{X} \text { is a prime number less than } 4) \\ & =\mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}=3) \\ & =0.23+0.10 \\ & =0.33 \\ \mathrm{P}(\mathrm{E} \cup \mathrm{F}) & =\mathrm{P}(\mathrm{E})+\mathrm{P}(\mathrm{F})-\mathrm{P}(\mathrm{E} \cap \mathrm{F}) \\ & =0.6+0.48-0.33 \\ & =0.75\end{aligned}\)