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:

$\sum\limits_0^3\text{P}(\text{x})=1$ $\text{k}+3\text{k}+3\text{k}+\text{k}=1$ $\text{k}=\frac{1}{8}$ $\text{x}$ $\text{P}(\text{x})$ $\text{x}\text{P}(\text{x})$ $\text{x}^2\text{P}(\text{x})$ $0$ $\frac{1}{8}$ $0$ $0$ $1$ $\frac{3}{8}$ $\frac{3}{8}$ $\frac{3}{8}$ $2$ $\frac{3}{8}$ $\frac{6}{8}$ $\frac{12}{8}$ $3$ $\frac{1}{8}$ $\frac{3}{8}$ $\frac{9}{8}$ Total $\text{E(x)}=\frac{12}{8}=1.5$ $\text{E}(\text{x}^2)=3$ $\text{V(x)}=\text{E}(\text{x}^2)-[\text{E}(\text{x})^2]$ $\text{V(x)}=3-(1.5)^2$ $\text{V(x)}=0.75=\frac{3}{4}$

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:

$\sum\limits_0^3\text{P}(\text{x})=1$ $\text{k}+3\text{k}+3\text{k}+\text{k}=1$ $\text{k}=\frac{1}{8}$ $\text{x}$ $\text{P}(\text{x})$ $\text{x}\text{P}(\text{x})$ $\text{x}^2\text{P}(\text{x})$ $0$ $\frac{1}{8}$ $0$ $0$ $1$ $\frac{3}{8}$ $\frac{3}{8}$ $\frac{3}{8}$ $2$ $\frac{3}{8}$ $\frac{6}{8}$ $\frac{12}{8}$ $3$ $\frac{1}{8}$ $\frac{3}{8}$ $\frac{9}{8}$ Total $\text{E(x)}=\frac{12}{8}=1.5$ $\text{E}(\text{x}^2)=3$ $\text{V(x)}=\text{E}(\text{x}^2)-[\text{E}(\text{x})^2]$ $\text{V(x)}=3-(1.5)^2$ $\text{V(x)}=0.75=\frac{3}{4}$

If X is a random variable with probability distribution as given below: X = x_i 0 1 2 3 P(X = X

M.C.Q (1 Marks)

GSEB - English Medium

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$\frac{1}{8},\frac{24}{27}$
D$\frac{1}{8},\frac{3}{4}$

STD 12 Science
Maths
Probability
Exemplar

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

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