A random variable X has the probability distr | vedclass

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(C) The given probability distribution for the random variable $X$ is:$\begin{array}{|c|c|c|c|c|c|c|}\hline X=x_i & 1 & 2 & 3 & 4 & 5 & 6 \\hline P(X

Vedclass · Accepted Answer

$0.82$

Vedclass · Answer

(C) The given probability distribution for the random variable $X$ is:$\begin{array}{|c|c|c|c|c|c|c|}\hline X=x_i & 1 & 2 & 3 & 4 & 5 & 6 \\hline P(X=x_i) & 0.2 & 0.3 & 0.12 & 0.1 & 0.2 & 0.08 \\hline \end{array}$We define the events $A$ and $B$ as follows:$A = \{x_i \mid x_i 	ext{ is a prime number}\} = \{2, 3, 5\}$$B = \{x_i \mid x_i The union of these two events is $A \cup B = \{1, 2, 3, 5\}$.Therefore,the probability $P(A \cup B)$ is the sum of the probabilities of these individual outcomes:$P(A \cup B) = P(X=1) + P(X=2) + P(X=3) + P(X=5)$$P(A \cup B) = 0.2 + 0.3 + 0.12 + 0.2 = 0.82$Thus,the correct option is $C$.

$X=x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X=x)$	$0.1$	$k$	$0.2$	$2k$	$0.3$	$k$

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