(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 $X$ is a random variable with the distribution given below:
$X = x_i$ $0$ $1$ $2$ $3$
$P(X = x_i)$ $k$ $3k$ $3k$ $k$

Then the value of $k$ and its variance are respectively given by:

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If $X$ is a Poisson variate such that $P(X=1) = 2P(X=2)$,then $P(X=3)$ is equal to:

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$A$ random variable $X$ has the probability distribution given below. Its variance is:
$X$ $1$ $2$ $3$ $4$ $5$
$P(X=x)$ $K$ $2K$ $3K$ $2K$ $K$

$X$	$1$	$2$	$3$	$4$	$5$
$P(X=x)$	$K$	$2K$	$3K$	$2K$	$K$

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$A$ random variable $X$ takes values $0, 1, 2, 3, \ldots$ with probability $P(X=x) = K(x+1)\left(\frac{1}{5}\right)^x$,where $K$ is a constant. Then $P(X=0)$ is:

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In a pizza hut,the following distribution is found for the daily demand of pizzas. Then the expected daily demand and variance are respectively:
No. of Pizzas $(x_i)$ $5$ $6$ $7$ $8$ $9$ $10$
Probability $(P_i)$ $0.07$ $0.2$ $0.3$ $0.3$ $0.07$ $0.06$

No. of Pizzas $(x_i)$	$5$	$6$	$7$	$8$	$9$	$10$
Probability $(P_i)$	$0.07$	$0.2$	$0.3$	$0.3$	$0.07$	$0.06$

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$X = x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$k$	$\frac{1}{5}$	$2k$	$\frac{3}{10}$	$k$

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 $X$ is a random variable with the distribution given below:
$X = x_i$ $0$ $1$ $2$ $3$
$P(X = x_i)$ $k$ $3k$ $3k$ $k$

Then the value of $k$ and its variance are respectively given by:

$A$ random variable $X$ has the probability distribution given below. Its variance is:
$X$ $1$ $2$ $3$ $4$ $5$
$P(X=x)$ $K$ $2K$ $3K$ $2K$ $K$

In a pizza hut,the following distribution is found for the daily demand of pizzas. Then the expected daily demand and variance are respectively:
No. of Pizzas $(x_i)$ $5$ $6$ $7$ $8$ $9$ $10$
Probability $(P_i)$ $0.07$ $0.2$ $0.3$ $0.3$ $0.07$ $0.06$