(C) Given the probability distribution:Mean $m = \sum p_i x_i = (0 \times \frac{1}{3}) + (1 \times \frac{1}{2}) + (2 \times 0) + (3 \times \frac{1}{6}

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(C) Given the probability distribution:Mean $m = \sum p_i x_i = (0 \times \frac{1}{3}) + (1 \times \frac{1}{2}) + (2 \times 0) + (3 \times \frac{1}{6})$$m = 0 + \frac{1}{2} + 0 + \frac{1}{2} = 1$Variance $\sigma^2 = \sum p_i (x_i - m)^2$$\sigma^2 = \frac{1}{3}(0 - 1)^2 + \frac{1}{2}(1 - 1)^2 + 0(2 - 1)^2 + \frac{1}{6}(3 - 1)^2$$\sigma^2 = \frac{1}{3}(1) + \frac{1}{2}(0) + 0 + \frac{1}{6}(4)$$\sigma^2 = \frac{1}{3} + 0 + 0 + \frac{4}{6} = \frac{1}{3} + \frac{2}{3} = 1$Thus,$m = 1$ and $\sigma^2 = 1$,so $m = \sigma^2 = 1$.

Class 12 Mathematics Probability Probability distribution

Medium

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:

$X=x$	$0$	$1$	$2$	$3$
$P(X=x)$	$\frac{1}{3}$	$\frac{1}{2}$	$0$	$\frac{1}{6}$

TS EAMCET 2009 All TS EAMCET

AP EAMCET 2009 All AP EAMCET

A
$m=\sigma^2=2$
B
$m=1, \sigma^2=2$
C
$m=\sigma^2=1$
D
$m=2, \sigma^2=1$

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Probability

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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 $P(2 < X < 6)$.

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

MediumMHT CET 2021

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The probability distribution of a random variable $X$ is given by
$X=x$ $0$ $1$ $2$ $3$ $4$
$P(X=x)$ $0.4$ $0.3$ $0.1$ $0.1$ $0.1$
The variance of $X$ is

EasyAP EAMCET 2007

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$A$ random variable $X$ has the following probability distribution:
$X = 1, 2, 3, 4, 5$
$P(X) = 0.1, 0.2, 0.3, 0.2, 0.2$
For the events $E = \{ X \text{ is a prime number} \}$ and $F = \{ X < 4 \}$, find $P(E \cap F)$.

DifficultMHT CET 2026

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$A$ box contains $8$ red and $N$ green balls. Two balls are drawn at random from it. If $X$ is the random variable representing the number of green balls drawn and $E(X) = 1.2$, then $N = ...$

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The discrete random variables $X$ and $Y$ are independent from one another and are defined as $X \sim B(16, 0.25)$ and $Y \sim P(2)$. Then the sum of the variances of $X$ and $Y$ is

MediumAP EAMCET 2022

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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:

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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 $P(2 < X < 6)$.

The probability distribution of a random variable $X$ is given by$X=x$$0$$1$$2$$3$$4$$P(X=x)$$0.4$$0.3$$0.1$$0.1$$0.1$The variance of $X$ is

$A$ random variable $X$ has the following probability distribution:$X = 1, 2, 3, 4, 5$$P(X) = 0.1, 0.2, 0.3, 0.2, 0.2$For the events $E = \{ X \text{ is a prime number} \}$ and $F = \{ X < 4 \}$, find $P(E \cap F)$.

$A$ box contains $8$ red and $N$ green balls. Two balls are drawn at random from it. If $X$ is the random variable representing the number of green balls drawn and $E(X) = 1.2$, then $N = ...$

The discrete random variables $X$ and $Y$ are independent from one another and are defined as $X \sim B(16, 0.25)$ and $Y \sim P(2)$. Then the sum of the variances of $X$ and $Y$ is

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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:

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 $P(2 < X < 6)$.

The probability distribution of a random variable $X$ is given by
$X=x$ $0$ $1$ $2$ $3$ $4$
$P(X=x)$ $0.4$ $0.3$ $0.1$ $0.1$ $0.1$
The variance of $X$ is

$A$ random variable $X$ has the following probability distribution:
$X = 1, 2, 3, 4, 5$
$P(X) = 0.1, 0.2, 0.3, 0.2, 0.2$
For the events $E = \{ X \text{ is a prime number} \}$ and $F = \{ X < 4 \}$, find $P(E \cap F)$.