Question

A random variable $X$ has the probability distribution $X$: $1,2,3,4,5,6,7,8$
$P\left(X\right):0.15,0.23,0.12,0.10,0.20,0.08,0.07,0.05$ . For the events $E=$ $\{$X is a prime number $\}$ and $F=\{X<4\}$, the probability $P(E\cup F)$ is:

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

Answer 1

The correct option is B $0.77$

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