Probability

ax² + bx + c = 0

D = b² – 4ac

D = 0

b² – 4ac = 0

b² = 4ac

(i) AC = 1, b = 2 (1, 2, 1) is one way

(ii) AC = 4, b = 4

$\begin{array}{l}a=4c=1\\ a=2c=2\\ a=1c=4\end{array}\}3ways$  

(iii) AC = 9, b = 6, a = 3, c = 3 is one way

1 + 3 + 1 = 5 way

Required probability =  $\frac{5}{216}$   

Let

A : Missile hit the target

B : Missile intercepted      

P (B) =    $\frac{1}{3}P\left(A/\overline{B}\right)=\frac{3}{4}$

$P\left(\overline{B}\right)=\frac{2}{3}$  

$\Rightarrow P\left(\overline{B}\cap A\right)=\frac{3}{4}*\frac{2}{3}=\frac{1}{2}$   

Required Probability = $\frac{2}{3}*\frac{3}{4}*\frac{2}{3}*\frac{3}{4}*\frac{2}{3}*\frac{3}{4}=\frac{1}{8}$  

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

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

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

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) = $\frac{13}{96}\left(given\right)$  

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

$x=\frac{1}{8}$