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 $F=\{X<4\}$ then $P(E\cup F)$ is

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

Answer 1

Answer: (B)

$E=\{$ is a prime number $\}=\{2,3,5,7\}$
$P(E)=P(X=2)+P(X=3)+P(X=5)+P(X=7)$
$P(E)=0.23+0.12+0.20+0.07=0.62$
$F=\{X<4\}=\{1,2,3\}$
$P(F)=P(X=1)=+P(X=2)+P(X=3)$
$P(F)=0.15+0.23+0.12=0.5$
$E\cap F=\{X$ is prime number as well as $<4\}=\{2,3\}$
$P(E\cap F)=P(X=2)+P(X=3)=0.23+0.12=0.35$
Therefore required probability
$P(E\cup F)=P(E)+P(F)-P(E\cap F)$
$\Rightarrow P(E\cup F)=0.62+0.5-0.35$
$\Rightarrow P(E\cup F)=0.77$