Probability

  $\overline{x}=\sum *p\left(x\right)$  

$\Rightarrow \frac{23}{10}=\frac{-2}{5}-a+1+\frac{4}{5}+6b$            

  $\Rightarrow 6b-a=\frac{9}{10}.........\left(i\right)$             

Also,   $\frac{1}{5}+a+\frac{1}{3}+\frac{1}{5}+b=1$

$\Rightarrow a+b=\frac{4}{15}.......\left(ii\right)$       

(i) & (ii) -> $a=\frac{1}{10},b=\frac{1}{6}$

$\Rightarrow 100{\sigma }^2=781$          

  $\underset{x\to \frac{\pi }{4}}{lim}\frac{\frac{\pi }{4}.f\left(se{c}^{2}x\right)2se{c}^{2}xtanx}{2x}$ = 2f (2)

Total ways = 6!

Ways satisfying g (3) = 2g (1) is 3

Number of onto function 3 * 4!

Probability = $\frac{1}{10}$

P (E₁)= 0.9 $P\left({\overline{E}}_t\right)=0.1andP\left(E_2\right)=0.8,P\left({\overline{E}}_2\right)=0.2$

$\therefore RequiredprobabilityP=\frac{0.8*0.1}{0.8*0.1+0.2*0.1+0.9*0.2}=\frac{0.8}{2.8}=\frac{2}{7}$

$\therefore 98P=28$

$\sum P\left(x\right)=1\Rightarrow k+2k+2k+3k+k=1sok=\frac{1}{9}$

 Now, $P\left(\frac{1<x<4}{x\angle 3}\right)=P\frac{\left(x=2\right)}{P\left(x\angle 3\right)}=\frac{\frac{2k}{9k}}{\frac{k}{9k}+\frac{2k}{9k}}=\frac{2}{3}$  

$\Rightarrow P=\frac{2}{3}$          

So, $5P=\lambda kgives\frac{10}{3}=\lambda *\frac{1}{9}\Rightarrow \lambda =30$  

Required probability = probability of both getting 0 head or 1 head or 2 head or 3 head

 = $\frac{5}{16}$

Required probability of obtaining a (sum = 7) = $\frac{13}{96}\left(given\right)$  

    $i.e.2\left(\left(\frac{1}{6}+x\right)\left(\frac{1}{6}-x\right)+\frac{1}{6}*\frac{1}{6}+\frac{1}{6}*\frac{1}{6}\right)=\frac{13}{96}$

$x=\frac{1}{8}$           

P – 2P² + 2P – 2P² = $\frac{5}{9}$  

$\Rightarrow P=\frac{5}{12},\frac{1}{3}$

89 – k + k + 98 – k + x = 100

=> $87\le k\le 89$

System of equations can be written as

  $\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 1\hfill & \hfill 3\hfill & \hfill \lambda \hfill \end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left(\begin{array}{l}5\\ \mu \\ 1\end{array}\right)$             

R₃ – R₂, R₂ – R₁

For unique solution $\lambda \ne 5,\mu \in R.P=\frac{5}{6}$  

For no solution $\lambda =5,\mu \ne 3q=\frac{1}{6}*\frac{5}{6}=\frac{5}{36}$