A random variable $x$ has the following probability distribution For the events $\mathrm{E}=\{\mathrm{X}$ is…
A random variable $x$ has the following probability distribution
For the events $\mathrm{E}=\{\mathrm{X}$ is prime number $\}$
$F=\{X \lt 4\}$
Then $P(E \cup F)=$
0.87
0.35
0.77
0.50
Solution
$\begin{aligned} & \mathrm{P}(\mathrm{E})=\mathrm{P}(\mathrm{X}=2 \text { or } \mathrm{X}=3 \text { or } \mathrm{X}=5 \text { or } \mathrm{X}=7) \\ &=\mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}=3)+\mathrm{P}(\mathrm{X}=5)+\mathrm{P}(\mathrm{X}=7) \\ &=0.23+0.12+0.20+0.07=0.62 \\ & \mathrm{P}(\mathrm{F})=\mathrm{P}(\mathrm{X} \lt 4) \\ &=\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}=3) \\ &=0.15+0.23+0.12=0.50 \\ & \mathrm{P}(\mathrm{E}\cap \mathrm{F}) \\ &=\mathrm{P}(\mathrm{X} \text { is a prime number less than } 4) \\ &=\mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}=3) \\ &=0.23+0.12=0.35 \\ & \therefore \quad \mathrm{P}(\mathrm{E} \cup \mathrm{F})=\mathrm{P}(\mathrm{E})+\mathrm{P}(\mathrm{F})-\mathrm{P}(\mathrm{E} \cap \mathrm{F}) \\ &=0.62+0.50-0.35=0.77\end{aligned}$
