For $\mathrm{k}=1,2,3$ the box $\mathrm{B}_{\mathrm{k}}$ contains k red balls and $(k+1)$ white balls. Let $P\left(B_1\right)=\frac{1}{2}, P\left(B_2\right)=\frac{1}{3}$ and $\mathrm{P}\left(\mathrm{B}_3\right)=\frac{1}{6} . \mathrm{A}$ box is selected at random and a ball is drawn from it. If a red ball is drawn from it, then the probability that it comes from box $\mathrm{B}_2$ is

A random variable $X$ takes the values $0,1,2,3$, $\qquad$ with probability

$\mathrm{P}(\mathrm{X}=x)=\mathrm{k}(x+1)\left(\frac{1}{5}\right)^x$, where k is a constant.

Then $\mathrm{P}(\mathrm{X}=0)$ is

If $u=\frac{\tan ^{-1} x}{\tan ^{-1} x+1}$ and $v=\tan ^{-1}\left(\tan ^{-1} x\right)$ then $\frac{d u}{d v}=$

If $\bar{a}=\frac{1}{\sqrt{10}}(3 \hat{i}+\hat{k})$ and $\bar{b}=\frac{1}{7}(2 \hat{i}+3 \hat{j}-6 \hat{k})$ then the value of $(2 \overline{\mathrm{a}}-\overline{\mathrm{b}}) \cdot[(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times(\overline{\mathrm{a}}+2 \overline{\mathrm{~b}})]=$