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(B) We know that the sum of probabilities in a probability distribution is $1$. Therefore,$\frac{25}{36} + k + \frac{1}{36} = 1 \Rightarrow k + \frac{

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$\frac{5}{18}$

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(B) We know that the sum of probabilities in a probability distribution is $1$. Therefore,$\frac{25}{36} + k + \frac{1}{36} = 1 \Rightarrow k + \frac{26}{36} = 1 \Rightarrow k = 1 - \frac{13}{18} = \frac{5}{18}$.The mean of a random variable $X$ is given by $E(X) = \sum x_i P(x_i)$.Given $E(X) = \frac{1}{3}$,we have:$E(X) = 0 	imes \frac{25}{36} + 1 	imes k + 2 	imes \frac{1}{36} = k + \frac{2}{36} = k + \frac{1}{18}$.Since $E(X) = \frac{1}{3}$,we have $k + \frac{1}{18} = \frac{1}{3} \Rightarrow k = \frac{1}{3} - \frac{1}{18} = \frac{6-1}{18} = \frac{5}{18}$.The variance of $X$ is given by $Var(X) = E(X^2) - [E(X)]^2$.First,calculate $E(X^2) = \sum x_i^2 P(x_i)$:$E(X^2) = 0^2 	imes \frac{25}{36} + 1^2 	imes k + 2^2 	imes \frac{1}{36} = 0 + k + \frac{4}{36} = k + \frac{1}{9}$.Substituting $k = \frac{5}{18}$:$E(X^2) = \frac{5}{18} + \frac{1}{9} = \frac{5+2}{18} = \frac{7}{18}$.Now,$Var(X) = \frac{7}{18} - (\frac{1}{3})^2 = \frac{7}{18} - \frac{1}{9} = \frac{7-2}{18} = \frac{5}{18}$.

$X$	$0$	$1$	$2$
$P(X)$	$\frac{25}{36}$	$k$	$\frac{1}{36}$

$A$ random variable $X$ has the following probability distribution:
$X$ $0$ $1$ $2$
$P(X)$ $\frac{25}{36}$ $k$ $\frac{1}{36}$

If the mean of the random variable $X$ is $\frac{1}{3}$,then the variance is:

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$A$ random variable $X$ has the following probability distribution:$X$$0$$1$$2$$P(X)$$\frac{25}{36}$$k$$\frac{1}{36}$If the mean of the random variable $X$ is $\frac{1}{3}$,then the variance is: