Probability

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 $\lambda =5,\mu \ne 3q=\frac{1}{6}*\frac{5}{6}=\frac{5}{36}$  

$P\left(\frac{x\ge 5}{x>2}\right)=\frac{P\left(x\ge 5\cap x>2\right)}{P\left(x>2\right)}=\frac{P\left(x\ge 5\right)}{P\left(x>2\right)}$

$=\frac{1-P\left(x\le 4\right)}{1-P\left(x\le 2\right)}$

$=\frac{1-\left[{\left(\frac{5}{6}\right)}^3*\frac{1}{6}+{\left(\frac{5}{6}\right)}^2*\frac{1}{6}+\left(\frac{5}{6}\right)\frac{1}{6}+\frac{1}{6}\right]}{1-\left[\frac{5}{6}*\frac{1}{6}+\frac{1}{6}\right]}$

$={\left(\frac{5}{6}\right)}^4*{\left(\frac{6}{5}\right)}^2={\left(\frac{5}{6}\right)}^2=\frac{25}{36}$

 $\begin{array}{ccc}\hfill 3R\hfill & \hfill 2R\hfill & \hfill 4B\hfill \\ \hfill 5B\hfill & \hfill 3W\hfill & \hfill 2W\hfill \end{array}$

I        II

Let

E₁ : a red ball is transferred from I to II

E₂ : a black is transferred from I to II

E₃ :a white transferred from I to II

E : a black ball is drawn from 2nd bag after a ball from I to II was transferred.

$P\left(\frac{E_1}{E}\right)=\frac{P\left(E_1\cap E\right)}{P\left(E\right)}$

$P\left(E\right)=P\left(E_1\cap E\right)+P\left(E_2\cap E\right)+P\left(E_3\cap E\right)$

= $P\left(E_1\right)\cdot P\left(\frac{E}{E_1}\right)+....+.....$

= $\frac{3}{10}\cdot \frac{5}{10}+\frac{4}{10}\cdot \frac{6}{10}+\frac{3}{10}\cdot \frac{5}{10}=\frac{54}{100}$

$P\left(E_1/E\right)=\frac{15/100}{54/100}=\frac{5}{18}$

x + y + z = 48

x, y, z  $\in$ {2, 4, 8, 16, 32, 32}

16 + 16 + 16 -> P_{1 $={\left(\frac{1}{16}\right)}^3$}  

 $32+8+8\to +P_2=\frac{1}{32}.{\left(\frac{1}{8}\right)}^2.\frac{3!}{2!}$

$P=\frac{1}{2^{12}}+\frac{3}{2^{11}}$

$=\frac{1+6}{2^{12}}=\frac{7}{2^{12}}$  

For $x^2+\alpha x=\beta >0\forall x\in R$  to hold, we should have   ${\alpha }^2-4\beta <0$

If $\alpha =1,\beta$  can be 1, 2, 3, 4, 5, 6 i.e., 6 choices

If a  = 4, b can be 5 or 6 i.e., 2 choices

hence total favourable outcomes

= 6 + 5 + 4 + 2 + 0 + 0 = 17

Required probability =  $\frac{17}{36}$

If n is number of trials, p is probability of success and q is probability of unsuccess then,

Mean = np and variance = npq.

Here np + npq = 24         …. (i)

np. npq = 128                   …. (ii)

and q = 1 – p                    …… (iii)

From eq. (i), (ii) and (iii) :

$=\left(32+\frac{32*31}{2}\right).\frac{1}{2^{32}}$      

$=\frac{33}{2^{28}}$

  $?\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$ …….(i)

then

$\overrightarrow{a}+\overrightarrow{c}=-\overrightarrow{b}$

then

$\left(\overrightarrow{a}+\overrightarrow{c}\right)*\overrightarrow{b}=-\overrightarrow{b}*\overline{b}$

$\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}$  $\therefore \overrightarrow{a}*\overrightarrow{b}+\overrightarrow{c}*\overrightarrow{b}=\overrightarrow{0}$ …….(ii)

Now (S1) : $\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}\text{exception 25:}$

$\left|\overrightarrow{a}*\overrightarrow{b}+\overrightarrow{c}*\overrightarrow{b}\right|-\left|\overrightarrow{c}\right|=6\left(2\sqrt[]{2}-1\right)$

$\therefore cos\left(\angle ACB\right)=\sqrt[]{\frac{2}{3}}$

$\therefore \angle ACB=co{s}^{-1}\sqrt[]{\frac{2}{3}}$

$$ $\therefore S\left(2\right)$  is correct

Find total (5 digit) number of divisors of 7. Find total (5 digit) number of divisors of 35