The range of a random variable X is {0, 1, 2} | vedclass

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(C) The sum of all probabilities for a random variable must be $1$. $P(X = 0) + P(X = 1) + P(X = 2) = 1$ $3c^3 + (4c - 10c^2) + (5c - 1) = 1$ $3c^3 -

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$\frac{8}{9}$

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(C) The sum of all probabilities for a random variable must be $1$. $P(X = 0) + P(X = 1) + P(X = 2) = 1$ $3c^3 + (4c - 10c^2) + (5c - 1) = 1$ $3c^3 - 10c^2 + 9c - 2 = 0$ Factoring the cubic equation,we get $(c - 1)(c - 2)(3c - 1) = 0$. Since $0 \le P(X) \le 1$,we test the values. If $c = 1$,$P(X = 2) = 5(1) - 1 = 4$,which is impossible. If $c = 2$,$P(X = 2) = 5(2) - 1 = 9$,which is impossible. Thus,$c = \frac{1}{3}$. Now,$P(X = 0) = 3(\frac{1}{3})^3 = \frac{1}{9}$,$P(X = 1) = 4(\frac{1}{3}) - 10(\frac{1}{3})^2 = \frac{4}{3} - \frac{10}{9} = \frac{2}{9}$,and $P(X = 2) = 5(\frac{1}{3}) - 1 = \frac{2}{3}$. We need to find $P(0 $P(0 < X \le 2) = \frac{2}{9} + \frac{2}{3} = \frac{2}{9} + \frac{6}{9} = \frac{8}{9}$.

$X = x$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X = x)$	$0.15$	$0.23$	$k$	$0.10$	$0.20$	$0.08$	$0.07$	$0.05$

The range of a random variable $X$ is $\{0, 1, 2\}$. If $P(X = 0) = 3c^3$,$P(X = 1) = 4c - 10c^2$,and $P(X = 2) = 5c - 1$,then find $P(0 < X \le 2)$.

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