- A$\frac{5}{108}$
- B$\frac{18}{108}$
- C$\frac{30}{108}$
- D$\frac{48}{108}$
Solution:
Total number of balls = 5brown + 4white = 9
Required probability $=\frac{5}{9}\times\frac{4}{8}+\frac{4}{9}\times\frac{3}{8}=\frac{4}{9}$
$\Rightarrow\ \frac{4\times12}{9\times12}=\frac{48}{108}$
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Statement $-I$ : Let $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$ and $\vec{b}=2 \hat{i}+\hat{j}-\hat{k}$. Then the vector $\vec{r}$ satisfying $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{r}}=0$ is of magnitude $\sqrt{10}$.
Statement $-II$ : In a triangle $A B C, \cos 2 A+\cos 2 B$ $+\cos 2 \mathrm{C} \geq-\frac{3}{2}$