| X = xi | 0 | 1 | 2 | 3 |
| P(X = Xi) | k | 3k | 3k | k |
The value of k and its variance are:
$\frac{1}{8},\frac{22}{27}$
$\frac{1}{8},\frac{23}{27}$
$\frac{1}{8},\frac{24}{27}$
$\frac{1}{8},\frac{3}{4}$
Solution:
$\sum\limits_0^3\text{P}(\text{x})=1$
$\text{k}+3\text{k}+3\text{k}+\text{k}=1$
$\text{k}=\frac{1}{8}$
| $\text{x}$ | $\text{P}(\text{x})$ | $\text{x}\text{P}(\text{x})$ | $\text{x}^2\text{P}(\text{x})$ |
| $0$ | $\frac{1}{8}$ | $0$ | $0$ |
| $1$ | $\frac{3}{8}$ | $\frac{3}{8}$ | $\frac{3}{8}$ |
| $2$ | $\frac{3}{8}$ | $\frac{6}{8}$ | $\frac{12}{8}$ |
| $3$ | $\frac{1}{8}$ | $\frac{3}{8}$ | $\frac{9}{8}$ |
| $\text{Total}$ |
| $\text{E(x)}=\frac{12}{8}=1.5$ | $\text{E}(\text{x}^2)=3$ |
$\text{V(x)}=\text{E}(\text{x}^2)-[\text{E}(\text{x})^2]$
$\text{V(x)}=3-(1.5)^2$
$\text{V(x)}=0.75=\frac{3}{4}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$P = \left\{ {\left( {a,b} \right):{{\sec }^2}\,a - {{\tan }^2}\,b = 1\,} \right\}$. Then $P$ is
$f\left( x \right) = \int_1^x {\left\{ {2\left( {t - 1} \right){{\left( {t - 2} \right)}^3} + 3{{\left( {t - 1} \right)}^2}{{\left( {t - 2} \right)}^2}} \right\}} dt$ is maximum when $x$ is equal to
$( S 1):|(\overrightarrow{ a } \times \overrightarrow{ b })+(\overrightarrow{ c } \times \overrightarrow{ b })|-|\overrightarrow{ c }|=6(2 \sqrt{2}-1)$
$( S 2): \angle ABC =\cos ^{-1}\left(\sqrt{\frac{2}{3}}\right)$. Then