(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:

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

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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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Three fair coins with faces numbered $1$ and $0$ are tossed simultaneously. The variance $(X)$ of the probability distribution of random variable $X$,where $X$ is the sum of numbers on the uppermost faces,is

MediumMHT CET 2023

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Which of the following cannot be a valid assignment of probabilities for outcomes of sample space $S = \{\omega_{1}, \omega_{2}, \omega_{3}, \omega_{4}, \omega_{5}, \omega_{6}, \omega_{7}\}$?

Outcome Probability

$\omega_{1}$ $0.1$

$\omega_{2}$ $0.2$

$\omega_{3}$ $0.3$

$\omega_{4}$ $0.4$

$\omega_{5}$ $0.5$

$\omega_{6}$ $0.6$

$\omega_{7}$ $0.7$

Medium

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The cumulative distribution function (c.d.f.) $F(x)$ associated with the probability density function (p.d.f.) $f(x) = 3(1 - x^2)$ for $0 < x < 1$ and $f(x) = 0$ otherwise,is given by $F(x) = k(x - \frac{2x^3}{k})$. Find the value of $k$.

EasyMHT CET 2020

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$A$ fair coin is tossed $2$ times. $A$ person receives $₹ X^{3}$ if he gets $X$ number of heads. His expected gain is $=$

MediumMHT CET 2020

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$A$ random variable $X$ has the following probability distribution:

$X$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$

$P(X)$ $0$ $k$ $2k$ $3k$ $3k^2$ $k^2$ $2k^2$ $7k^2+k$

Determine $P(0 < X < 3)$. (in $/10$)

Easy

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Outcome	Probability
$\omega_{1}$	$0.1$
$\omega_{2}$	$0.2$
$\omega_{3}$	$0.3$
$\omega_{4}$	$0.4$
$\omega_{5}$	$0.5$
$\omega_{6}$	$0.6$
$\omega_{7}$	$0.7$

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

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:

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Three fair coins with faces numbered $1$ and $0$ are tossed simultaneously. The variance $(X)$ of the probability distribution of random variable $X$,where $X$ is the sum of numbers on the uppermost faces,is

The cumulative distribution function (c.d.f.) $F(x)$ associated with the probability density function (p.d.f.) $f(x) = 3(1 - x^2)$ for $0 < x < 1$ and $f(x) = 0$ otherwise,is given by $F(x) = k(x - \frac{2x^3}{k})$. Find the value of $k$.

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$A$ random variable $X$ has the following probability distribution: $X$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $P(X)$ $0$ $k$ $2k$ $3k$ $3k^2$ $k^2$ $2k^2$ $7k^2+k$ Determine $P(0 < X < 3)$. (in $/10$)

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

Then:

$A$ random variable $X$ has the following probability distribution:

$X$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$

$P(X)$ $0$ $k$ $2k$ $3k$ $3k^2$ $k^2$ $2k^2$ $7k^2+k$

Determine $P(0 < X < 3)$. (in $/10$)