A random variable \(\mathrm{X}\) has the probability distribution

For the events \(\mathrm{E}=\{\mathrm{X}\) is a prime number \(\}\) and \(\mathrm{F}=\{x < 5\}, \mathrm{P}(\mathrm{E} \cup \mathrm{F})\) is

\(\begin{aligned} \mathrm{P}(\mathrm{E})= & \mathrm{P}(\mathrm{X}=2 \text { or } \mathrm{X}=3 \text { or } \mathrm{X}=5 \text { or } \mathrm{X}=7) \\ = & \mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}=3)+\mathrm{P}(\mathrm{X}=5) \\ & +\mathrm{P}(\mathrm{X}=7)\end{aligned}\)
\(\begin{aligned} & =0.15+0.23+0.12+0.20 \\ & =0.7\end{aligned}\)
\(\begin{aligned} \mathrm{P}(\mathrm{E} \cap \mathrm{F}) & =\mathrm{P}(\mathrm{X} \text { is a prime number less than } 5) \\ & =\mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}=3) \\ & =0.23+0.12 \\ & =0.35 \\ \mathrm{P}(\mathrm{E} \cup \mathrm{F}) & =\mathrm{P}(\mathrm{E})+\mathrm{P}(\mathrm{F})-\mathrm{P}(\mathrm{E} \cap \mathrm{F}) \\ & =0.48+0.7-0.35 \\ & =0.83\end{aligned}\)