A random variable $\mathrm{X}$ has the probability distribution $ \begin{array}{|l|c|c|c|c|c|c|c|c|} \hline…
A random variable $\mathrm{X}$ has the probability distribution
$
\begin{array}{|l|c|c|c|c|c|c|c|c|}
\hline \mathrm{X}=x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline \mathrm{P}(\mathrm{X}=x) & 0.15 & 0.23 & 0.12 & 0.20 & 0.08 & 0.10 & 0.05 & 0.07 \\\hline\end{array}$
For the events $\mathrm{E}=\{\mathrm{X}$ is a prime number $\}$ and $\mathrm{F}=\{x < 5\}, \mathrm{P}(\mathrm{E} \cup \mathrm{F})$ is
0.63
0.75
0.83
0.90
Solution
$\begin{aligned} \mathrm{P}(\mathrm{E})= & \mathrm{P}(\mathrm{X}=2 \text { or } \mathrm{X}=3 \text { or } \mathrm{X}=5 \text { or } \mathrm{X}=7) \\ = & \mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}=3)+\mathrm{P}(\mathrm{X}=5) \\ & +\mathrm{P}(\mathrm{X}=7)\end{aligned}$
$\begin{aligned} & =0.15+0.23+0.12+0.20 \\ & =0.7\end{aligned}$
$\begin{aligned} \mathrm{P}(\mathrm{E} \cap \mathrm{F}) & =\mathrm{P}(\mathrm{X} \text { is a prime number less than } 5) \\ & =\mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}=3) \\ & =0.23+0.12 \\ & =0.35 \\ \mathrm{P}(\mathrm{E} \cup \mathrm{F}) & =\mathrm{P}(\mathrm{E})+\mathrm{P}(\mathrm{F})-\mathrm{P}(\mathrm{E} \cap \mathrm{F}) \\ & =0.48+0.7-0.35 \\ & =0.83\end{aligned}$