Question

A random variable $x$ has the probability distribution.

$x$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$p\left(x\right)$	$0.15$	$0.23$	$0.12$	$0.10$	$0.20$	$0.08$	$0.07$	$0.05$

If the events $E=${$x$ is prime number} and $F=\left\{x<4\right\}$, then $P\left(EUF\right)$ is equal to

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

Answer 1

The correct option is A

$0.77$

Explanation for the correct option:

Step 1: Find the probability of Event E and F:

Sum up the probability of all prime numbers to get the probability of Event E.

$\begin{array}{rcl}P\left(E\right)& =& P\left(2\right)+P\left(3\right)+P\left(5\right)+P\left(7\right)\\ & =& 0.23+0.12+0.20+0.07\\ & =& 0.62\end{array}$

Sum up the probability of all Event $1,2,3$ to get the probability of Event F.

$\begin{array}{rcl}P\left(F\right)& =& P\left(1\right)+P\left(2\right)+P\left(3\right)\\ & =& 0.15+0.23+0.12\\ & =& 0.5\end{array}$

Step 2: Find the value of $P\left(E\cup F\right)$:

Find the intersection of both events:

$\begin{array}{rcl}P\left(E\cap F\right)& =& P(x=2)+P(x=3)\\ & =& 0.23+0.12\\ & =& 0.35\end{array}$

Find the value of $P\left(E\cup F\right)$.

$\begin{array}{rcl}P\left(E\cup F\right)& =& P\left(E\right)+P\left(F\right)-P(E\cap F)\\ & =& 0.62+0.5-0.35\\ & =& 0.77\end{array}$

Hence, option (A) is the correct answer.