(A) For a probability distribution,the sum of all probabilities must be equal to $1$.Therefore,$\sum P(X = x) = 1$.$\frac{1}{10} + k + \frac{1}{5} + 2

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(A) For a probability distribution,the sum of all probabilities must be equal to $1$.Therefore,$\sum P(X = x) = 1$.$\frac{1}{10} + k + \frac{1}{5} + 2k + \frac{3}{10} + k = 1$Combine the constant terms and the terms with $k$:$(\frac{1}{10} + \frac{2}{10} + \frac{3}{10}) + (k + 2k + k) = 1$$\frac{6}{10} + 4k = 1$$4k = 1 - \frac{6}{10}$$4k = \frac{10 - 6}{10}$$4k = \frac{4}{10}$$k = \frac{1}{10}$

Class 12 Mathematics Probability Probability distribution

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The distribution of a random variable $X$ is given below. The value of $k$ is:
$X = x$ $-2$ $-1$ $0$ $1$ $2$ $3$
$P(X = x)$ $\frac{1}{10}$ $k$ $\frac{1}{5}$ $2k$ $\frac{3}{10}$ $k$

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A
$\frac{1}{10}$
B
$\frac{2}{10}$
C
$\frac{3}{10}$
D
$\frac{7}{10}$

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If $m$ and $\sigma^2$ are the mean and variance of the random variable $X$,whose distribution is given by:
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For the probability distribution of $X$ given below,the variance of $X$ is:
$X = x$ $-2$ $-1$ $0$ $1$ $2$
$P(X = x)$ $0.2$ $0.3$ $0.1$ $0.15$ $0.25$

$X = x$	$-2$	$-1$	$0$	$1$	$2$
$P(X = x)$	$0.2$	$0.3$	$0.1$	$0.15$	$0.25$

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If a random variable $x$ has the probability distribution as follows:
$x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
$P(x)$ $0$ $2k$ $k$ $3k$ $2k^2$ $2k$ $k^2+k$ $7k^2$

Then $P(3 < x \leq 6)$ is equal to:

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For a random variable $X$,if $P(X=k) = \frac{(k+1)a}{3^k}$ for $k=0, 1, 2, \ldots$,then $a = $

MediumAP EAMCET 2017

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$x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(x)$	$0$	$2k$	$k$	$3k$	$2k^2$	$2k$	$k^2+k$	$7k^2$

The distribution of a random variable $X$ is given below. The value of $k$ is:$X = x$$-2$$-1$$0$$1$$2$$3$$P(X = x)$$\frac{1}{10}$$k$$\frac{1}{5}$$2k$$\frac{3}{10}$$k$

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The distribution of a random variable $X$ is given below. The value of $k$ is:
$X = x$ $-2$ $-1$ $0$ $1$ $2$ $3$
$P(X = x)$ $\frac{1}{10}$ $k$ $\frac{1}{5}$ $2k$ $\frac{3}{10}$ $k$

If $m$ and $\sigma^2$ are the mean and variance of the random variable $X$,whose distribution is given by:
$X=x$ $0$ $1$ $2$ $3$
$P(X=x)$ $\frac{1}{3}$ $\frac{1}{2}$ $0$ $\frac{1}{6}$

Then:

For the probability distribution of $X$ given below,the variance of $X$ is:
$X = x$ $-2$ $-1$ $0$ $1$ $2$
$P(X = x)$ $0.2$ $0.3$ $0.1$ $0.15$ $0.25$

If a random variable $x$ has the probability distribution as follows:
$x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
$P(x)$ $0$ $2k$ $k$ $3k$ $2k^2$ $2k$ $k^2+k$ $7k^2$

Then $P(3 < x \leq 6)$ is equal to: