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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,$Var(X) = \frac{7}{18} - (\frac{1}{3})^2 = \frac{7}{18} - \frac{1}{9} = \frac{7-2}{18} = \frac{5}{18}$.

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

$X$	$1$	$2$	$3$	$4$	$5$	$6$
$P(X)$	$K$	$2K$	$3K$	$4K$	$5K$	$6K$

$Z$	$3$	$2$	$1$	$0$	$-1$
$P(Z)$	$0.3$	$0.2$	$0.4$	$0.1$	$0.05$

$A$ 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:

Similar Questions

The probability distribution of a discrete random variable $X$ is given by the following table:
$X$ $1$ $2$ $3$ $4$ $5$ $6$
$P(X)$ $K$ $2K$ $3K$ $4K$ $5K$ $6K$

Find the value of $P(2 < X < 6)$.

$A$ person throws an unbiased die. If the number shown is even,he gains an amount equal to the number shown. If the number is odd,he loses an amount equal to the number shown. Then his expectation is ₹.

State which of the following is not the probability distribution of a random variable. Give reasons for your answer.

$Z$ $3$ $2$ $1$ $0$ $-1$

$P(Z)$ $0.3$ $0.2$ $0.4$ $0.1$ $0.05$

Given the probability density function (p.d.f.) of the random variable $X$ as $f(x) = \frac{1}{2a}$ for $0 < x < 2a$ and $f(x) = 0$ otherwise, where $a > 0$, which of the following is correct?

Let a sample space be $S = \{\omega_{1}, \omega_{2}, \ldots, \omega_{6}\}$. Which of the following assignments of probabilities to each outcome is valid?

Outcome $\omega_1$ $\omega_2$ $\omega_3$ $\omega_4$ $\omega_5$ $\omega_6$

$(b)$ $1$ $0$ $0$ $0$ $0$ $0$

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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: