(B) Step $1$: Calculate the Expected value $E(X) = \sum x_i P(x_i)$.$E(X) = (1 \times 1/5) + (2 \times 2/5) + (3 \times 2/5) = 1/5 + 4/5 + 6/5 = 11/5$

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(B) Step $1$: Calculate the Expected value $E(X) = \sum x_i P(x_i)$.$E(X) = (1 \times 1/5) + (2 \times 2/5) + (3 \times 2/5) = 1/5 + 4/5 + 6/5 = 11/5$.Step $2$: Calculate $E(X^2) = \sum x_i^2 P(x_i)$.$E(X^2) = (1^2 \times 1/5) + (2^2 \times 2/5) + (3^2 \times 2/5) = 1/5 + 8/5 + 18/5 = 27/5$.Step $3$: Calculate Variance $Var(X) = E(X^2) - [E(X)]^2$.$Var(X) = 27/5 - (11/5)^2 = 27/5 - 121/25 = (135 - 121) / 25 = 14/25$.Thus, the values are $11/5$ and $14/25$.

Class 12 Mathematics Probability Probability distribution

Difficult

For the following probability distribution of a random variable $X$, the Expected value $E(X)$ and Variance $Var(X)$ of $X$ are respectively:
$X=x$ $1$ $2$ $3$
$P(X=x)$ $1/5$ $2/5$ $2/5$

MHT CET 2026 All MHT CET

A
$27/5, 27/25$
B
$11/5, 14/25$
C
$4/5, 14/25$
D
$7/5, 11/25$

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If $X$ is a random variable with cumulative distribution function $F(x)$ and its probability distribution is given by the following table:
$X = x$ $-1.5$ $-0.5$ $0.5$ $1.5$ $2.5$
$P(X = x)$ $0.05$ $0.2$ $0.15$ $0.25$ $0.35$

Then,find the value of $F(1.5) - F(-0.5)$.

$X = x$	$-1.5$	$-0.5$	$0.5$	$1.5$	$2.5$
$P(X = x)$	$0.05$	$0.2$	$0.15$	$0.25$	$0.35$

MediumMHT CET 2020

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The p.m.f. of a random variable $X$ is $P(x) = \begin{cases} \frac{2x}{n(n+1)}, & x = 1, 2, 3, \ldots, n \\ 0, & \text{otherwise} \end{cases}$,then $E(X)$ is

MediumMHT CET 2023

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For a Poisson variate $X$,if $P(X=2)=3 P(X=3)$,then the mean of $X$ is:

DifficultAP EAMCET 2003

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The cumulative distribution function $F(X)$ of a discrete random variable $X$ is given by the following table:
$X$ $1$ $2$ $3$ $4$ $5$ $6$
$F(X=x)$ $0.2$ $0.37$ $0.48$ $0.62$ $0.85$ $1$

Then $P[X=4] + P[X=5] = $

$X$	$1$	$2$	$3$	$4$	$5$	$6$
$F(X=x)$	$0.2$	$0.37$	$0.48$	$0.62$	$0.85$	$1$

MediumMHT CET 2021

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The probability distribution of a discrete random variable $X$ is given below:
$X = x$ $-1$ $0$ $1$ $2$
$P(X = x)$ $\frac{1}{3}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{3}$

Then the value of $6 \Sigma(x^2) P(X=x) - \operatorname{var}(X) =$ ?

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

MediumAP EAMCET 2025

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For the following probability distribution of a random variable $X$, the Expected value $E(X)$ and Variance $Var(X)$ of $X$ are respectively: $X=x$$1$$2$$3$$P(X=x)$$1/5$$2/5$$2/5$

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If $X$ is a random variable with cumulative distribution function $F(x)$ and its probability distribution is given by the following table:$X = x$$-1.5$$-0.5$$0.5$$1.5$$2.5$$P(X = x)$$0.05$$0.2$$0.15$$0.25$$0.35$Then,find the value of $F(1.5) - F(-0.5)$.

The p.m.f. of a random variable $X$ is $P(x) = \begin{cases} \frac{2x}{n(n+1)}, & x = 1, 2, 3, \ldots, n \\ 0, & \text{otherwise} \end{cases}$,then $E(X)$ is

For a Poisson variate $X$,if $P(X=2)=3 P(X=3)$,then the mean of $X$ is:

The cumulative distribution function $F(X)$ of a discrete random variable $X$ is given by the following table:$X$$1$$2$$3$$4$$5$$6$$F(X=x)$$0.2$$0.37$$0.48$$0.62$$0.85$$1$Then $P[X=4] + P[X=5] = $

The probability distribution of a discrete random variable $X$ is given below: $X = x$$-1$$0$$1$$2$$P(X = x)$$\frac{1}{3}$$\frac{1}{6}$$\frac{1}{6}$$\frac{1}{3}$Then the value of $6 \Sigma(x^2) P(X=x) - \operatorname{var}(X) =$ ?

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For the following probability distribution of a random variable $X$, the Expected value $E(X)$ and Variance $Var(X)$ of $X$ are respectively:
$X=x$ $1$ $2$ $3$
$P(X=x)$ $1/5$ $2/5$ $2/5$

If $X$ is a random variable with cumulative distribution function $F(x)$ and its probability distribution is given by the following table:
$X = x$ $-1.5$ $-0.5$ $0.5$ $1.5$ $2.5$
$P(X = x)$ $0.05$ $0.2$ $0.15$ $0.25$ $0.35$

Then,find the value of $F(1.5) - F(-0.5)$.

The cumulative distribution function $F(X)$ of a discrete random variable $X$ is given by the following table:
$X$ $1$ $2$ $3$ $4$ $5$ $6$
$F(X=x)$ $0.2$ $0.37$ $0.48$ $0.62$ $0.85$ $1$

Then $P[X=4] + P[X=5] = $

The probability distribution of a discrete random variable $X$ is given below:
$X = x$ $-1$ $0$ $1$ $2$
$P(X = x)$ $\frac{1}{3}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{3}$

Then the value of $6 \Sigma(x^2) P(X=x) - \operatorname{var}(X) =$ ?