Question

A random variable $X$ has the probability distribution:

$X$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X)$	$0.15$	$0.23$	$0.12$	$0.10$	$0.20$	$0.08$	$0.07$	$0.05$

For the events $E=\{X$ is a prime number $\}$ and $F=\{X<4\}$, the probability $P(E\cup F)$ is

• A. 0.50
• B. 0.77
• C. 0.35
• D. 0.87

Answer 1

Answer: (B)

$P(E)=P$ ($2$ or $3$ or $5$ or $7$)
$=0.23+0.12+0.20+0.07=0.62$
$P(F)=P$ ($1$ or $2$ or $3$)
$=0.15+0.23+0.12=0.50$
$P(E\cap F)=P$ ($2$ or $3$)
$=0.23+0.12=0.35$
$\therefore P(EUF)=P(E)+P(F)-P(E\cap F)$
$=0.62+0.50-0.35=0.77$