The range of a discrete random variable X is | vedclass

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(A) For a discrete random variable, the sum of all probabilities must be equal to $1$. Therefore, $P(X=1) + P(X=2) + P(X=3) = 1$. Substituting the giv

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$\frac{2}{3}$

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(A) For a discrete random variable, the sum of all probabilities must be equal to $1$. Therefore, $P(X=1) + P(X=2) + P(X=3) = 1$. Substituting the given values: $3k^3 + 2k^2 + (7 - 19k) = 1$. $3k^3 + 2k^2 - 19k + 6 = 0$. By testing values, if $k = 3$, $3(27) + 2(9) - 19(3) + 6 = 81 + 18 - 57 + 6 = 48 
eq 0$. If $k = \frac{1}{3}$, $3(\frac{1}{27}) + 2(\frac{1}{9}) - 19(\frac{1}{3}) + 6 = \frac{1}{9} + \frac{2}{9} - \frac{57}{9} + \frac{54}{9} = 0$. So, $k = \frac{1}{3}$ is a root. Now, $P(X=3) = 7 - 19k = 7 - 19(\frac{1}{3}) = 7 - \frac{19}{3} = \frac{21 - 19}{3} = \frac{2}{3}$.

$X=x$	$0$	$1$	$2$	$3$	$4$
$P(X=x)$	$0.4$	$0.3$	$0.1$	$0.1$	$0.1$

$X=x$	$1$	$2$	$3$	$4$
$P(X=x)$	$0.1$	$0.2$	$0.3$	$0.4$

$X = x$	$1$	$2$	$3$	$4$
$P(X = x)$	$1/10$	$2/10$	$3/10$	$4/10$

The range of a discrete random variable $X$ is $\{1, 2, 3\}$ and the probabilities of its elements are given by $P(X=1) = 3k^3$, $P(X=2) = 2k^2$, and $P(X=3) = 7 - 19k$. Then $P(X=3) = $

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