If X is a random variable with probability distribution as given … - Vidyadip

MCQ

If $X$ is a random variable with probability distribution as given below:

$X = x_i$	$0$	$1$	$2$	$3$
$P(X = X_i)$	$k$	$3k$	$3k$	$k$

The value of $k$ and its variance are:

A
$\frac{1}{8},\frac{22}{27}$
B
$\frac{1}{8},\frac{23}{27}$
C
$\frac{1}{8},\frac{24}{27}$
✓
$\frac{1}{8},\frac{3}{4}$

✓

Answer

Correct option: D.

$\frac{1}{8},\frac{3}{4}$

$\sum\limits_0^3\text{P}(\text{x})=1$
$\text{k}+3\text{k}+3\text{k}+\text{k}=1$
$\text{k}=\frac{1}{8}$

$\text{x}$	$\text{P}(\text{x})$	$\text{x}\text{P}(\text{x})$	$\text{x}^2\text{P}(\text{x})$
$0$	$\frac{1}{8}$	$0$	$0$
$1$	$\frac{3}{8}$	$\frac{3}{8}$	$\frac{3}{8}$
$2$	$\frac{3}{8}$	$\frac{6}{8}$	$\frac{12}{8}$
$3$	$\frac{1}{8}$	$\frac{3}{8}$	$\frac{9}{8}$
$\text{Total}$		$\text{E(x)}=\frac{12}{8}=1.5$	$\text{E}(\text{x}^2)=3$

$\text{V(x)}=\text{E}(\text{x}^2)-[\text{E}(\text{x})^2]$
$\text{V(x)}=3-(1.5)^2$
$\text{V(x)}=0.75=\frac{3}{4}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from Probability→Probability — all question types→MATHS STD 12 Science question papers→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

If $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ are three non-zero vectors, no two of which are collinear and the vector $\vec{\text{a}}+\vec{\text{b}}$ is collinear with $\vec{\text{c}}$, $\vec{\text{b}}+\vec{\text{c}}$ is collinear with $\vec{\text{a}}$, then $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=$

$\vec{\text{a}}$
$\vec{\text{b}}$
$\vec{\text{c}}$
None of these

The solution of $\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}}=\frac{1}{\sqrt{1+\text{x}^2}}$ is:

$\text{y}=\frac{1+\text{x}^2}{\text{x}}+\frac{\text{c}}{\text{x}}$
$\text{y}=\frac{\sqrt{1+\text{x}^2}}{\text{x}}+\frac{\text{c}}{\text{x}}$
$\text{y}=\frac{\text{x}}{\sqrt{1+\text{x}^2}}+\text{cx}$
$\text{None of these}$

The area bounded by the parabola $x = 4 - y^2$ and $y-$axis, in square units, is:

The set of points where the function $f(x) = x |x|$ is differentiable is:

The area of the ellipse $\frac{\text{x}2}{9}+\frac{\text{y}^2}{4}=1$ in first quadrant is $6\pi$ sq. units.
The ellipse is rotated about its centre in anti-clockwise direction till its major axis coincides with y-axis. Now the area of the ellipse in first Quadrant is $\pi$ sq. units.

2
4
6
8

If $A=\{1,2,3\}$ then the number of equivalence relations with element (1, 2) is-

If the constraints in a linear programming problem are changed

Let X represent the difference between the number of heads and the number of tails obtained when a coin is tossed 6 times. What are possible values of X?

Evaluate: $\int\frac{1}{\sqrt{9+8\text{x}-\text{x}^2}}\text{dx}.$

$-\sin^{-1}(\frac{\text{x}-4}{5})+\text{c}$
$\sin^{-1}(\frac{\text{x}+4}{5})+\text{c}$
$\sin^{-1}(\frac{\text{x}-4}{5})+\text{c}$
$\text{None of there}$

If $\int\frac{1}{5+4\sin\text{x}}\text{ dx}=\text{A}\tan^{-1}\Big(\text{B}\tan\frac{\pi}{2}+\frac{4}{3}\Big)+\text{C},$ then:

$\text{A}=\frac{2}{3},\text{B}=\frac{5}{3}$
$\text{A}=\frac{1}{3},\text{B}=\frac{2}{3}$
$\text{A}=-\frac{2}{3},\text{B}=\frac{5}{3}$
$\text{A}=\frac{1}{3},\text{B}=-\frac{5}{3}$