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Probability Distribution

The probabili...

Question

The probability mass function (p.m.f) of $X$ is given below:
$X = x$ $1$ $2$ $3$
$P (X = x)$ $\frac{1}{5}$ $\frac{2}{5}$ $\frac{2}{5}$

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Solution

We know, $E (X^{2}) = \sum X^{2} P (X)$
Therefore, $E (X^{2}) = 1^{2} \times \frac{1}{5} + 2^{2} \times \frac{2}{5} + 3^{2} \times \frac{2}{5}$
$= \frac{1}{5} + \frac{8}{5} + \frac{18}{5} = \frac{27}{5}$

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Similar questions

X

Y

$X$	$0$	$1$	$2$	3
$P (X)$	$\frac{1}{5}$	$\frac{2}{5}$	$\frac{1}{5}$	$\frac{1}{5}$

$Y$	$0$	$1$	$2$	$3$
$P (Y)$	$\frac{1}{5}$	$\frac{3}{10}$	$\frac{2}{5}$	$\frac{1}{10}$

X

$x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$
$P (X = x)$	$k$	$3 k$	$5 k$	$7 k$	$9 k$	$11 k$	$13 k$

k

P (X < 4), P (X

\geq

5)

P (3 < X

\leq

6)

x

P (X

\leq

x) >

\frac{1}{2}

$X = x$	0	1	2	3	4	5	6	7
$P (X = x)$	$0$	$K$	$2 K$	$2 k$	$3 K$	$K^{2}$	$2 K^{2}$	$7 K^{2} + k$

P (0 < X < 5) =

X	1	2	3	4	5	6
P(X=x)	k	2k	3k	4k	5k	6k

X

P (X \leq 4)

X

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

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