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
$P ( E \cup F )= P ( E )+ P ( F )- P ( E \cap F )$
$\Rightarrow P ( E \cup F )=0.62+0.5-0.35$
$\Rightarrow P ( E \cup F )=0.77$