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

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(B) For a probability distribution,the sum of all probabilities must be equal to $1$. $P(X=0) + P(X=1) + P(X=2) = 1$ $3C^3 + (4C - 10C^2) + (5C - 1) =

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$\frac{1}{3}$

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(B) For a probability distribution,the sum of all probabilities must be equal to $1$. $P(X=0) + P(X=1) + P(X=2) = 1$ $3C^3 + (4C - 10C^2) + (5C - 1) = 1$ $3C^3 - 10C^2 + 9C - 2 = 0$ By testing values,we find that $(C - 1)$ is a factor. Dividing the polynomial,we get $(C - 1)(3C^2 - 7C + 2) = 0$. Factoring further: $(C - 1)(3C - 1)(C - 2) = 0$. Thus,the possible values for $C$ are $1, \frac{1}{3}, 2$. We must check if these values result in probabilities between $0$ and $1$: If $C = 2$,$P(X=2) = 5(2) - 1 = 9$,which is $> 1$. So $C 
eq 2$. If $C = 1$,$P(X=1) = 4(1) - 10(1)^2 = -6$,which is $If $C = \frac{1}{3}$,$P(X=0) = 3(\frac{1}{27}) = \frac{1}{9}$,$P(X=1) = 4(\frac{1}{3}) - 10(\frac{1}{9}) = \frac{12-10}{9} = \frac{2}{9}$,and $P(X=2) = 5(\frac{1}{3}) - 1 = \frac{2}{3}$. Since all probabilities are between $0$ and $1$ and sum to $1$,the correct value is $C = \frac{1}{3}$.

$X = x$	$1$	$2$	$3$	$\dots$	$n$
$P(X = x)$	$\frac{1}{n}$	$\frac{1}{n}$	$\frac{1}{n}$	$\dots$	$\frac{1}{n}$

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 the value of $C$ is

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