Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

$\left(p\vee q\right)\Rightarrow q$

= $\sim \left(p\vee q\right)\vee q$

= $\left(\sim p\wedge \sim q\right)\vee q$

= $\left(\sim p\vee q\right)\wedge \left(\sim q\vee q\right)$

= $\sim p\vee q$

now $\left(p\wedge q\right)\Delta \left(\sim p\vee q\right)istautolog\text{?}y$

pq $p\wedge q$ $\sim p\vee q$ $\left(p\wedge q\right)\Delta \left(\sim p\vee q\right)$

TTTTT

TFFFT

FTFTT

FFFTT

So,  $\Delta =\Rightarrow$

Mean = 4 = $\mu$ = np

Variance = ${\sigma }^2=np\left(1-p\right)=\frac{4}{3}\Rightarrow 4\left(1-p\right)=\frac{4}{3}\Rightarrow p=\frac{2}{3}\Rightarrow n=6$

${=}^6{C}_0{\left(\frac{1}{3}\right)}^6{+}^6{C}_1{\left(\frac{2}{3}\right)}^1{\left(\frac{1}{3}\right)}^6{+}^6{C}_2{\left(\frac{2}{3}\right)}^2{\left(\frac{1}{3}\right)}^4=\frac{146}{27}$

X  0 1 2 3

P (X) $\frac{1}{6}$ $\frac{1}{2}$ $\frac{3}{10}$ $\frac{1}{30}$

${\sigma }^2=\Sigma x^{2}P\left(x\right)-\left(\Sigma *P{\left(x\right)}^2=\frac{56}{100}\right)$

$100{\sigma }^2=56$

 $P\left(A/B\right)=\frac{1}{7}\Rightarrow \frac{P\left(A\cap B\right)}{P\left(B\right)}=\frac{1}{7}$

$\Rightarrow P\left(B\right)=\frac{7}{9}$

$P\left(B/A\right)=\frac{2}{5}\Rightarrow \frac{P\left(A\cap B\right)}{P\left(A\right)}=\frac{2}{5}$

$\Rightarrow P\left(A\right)=\frac{5}{18}$

Now,  $P\left(A\text{'}\cup B\right)=1-P\left(A\cup B\right)+P\left(B\right)$

Both (S1 ) and (S2 ) are true

We have,  $1-$  (probability of all shots result in failure)  $>\frac{1}{4}$

$\begin{array}{rr}& \Rightarrow 1-{\left(\frac{9}{10}\right)}^{\mathrm{n}}>\frac{1}{4}\\ & \Rightarrow \frac{3}{4}>{\left(\frac{9}{10}\right)}^{\mathrm{n}}\Rightarrow \mathrm{n}\ge 3\end{array}$

$y+2x=\sqrt[]{11}+7\sqrt[]{7}$ ……… (i)

$2y+x=2\sqrt[]{11}+6\sqrt[]{7}$ ……… (ii)

$\Rightarrow x+y=\sqrt[]{11}+\frac{13}{3}\sqrt[]{7}$ ……… (iii)

Centre of the circle given by solving (i) & (ii)

$as\left(\frac{8\sqrt[]{7}}{3},\sqrt[]{11}+\frac{5\sqrt[]{7}}{3}\right)$

Again $\sqrt[]{11}y-3x=\frac{5\sqrt[]{77}}{3}$ is tangent to the circle.

$\therefore r=\left|\frac{\sqrt[]{11}\left(\sqrt[]{11}+\frac{5\sqrt[]{7}}{3}\right)-8\sqrt[]{7}-\frac{5\sqrt[]{77}}{3}}{\sqrt[]{11+9}}\right|=\left|\frac{11-8\sqrt[]{7}}{\sqrt[]{20}}\right|$

$\therefore {\left(5h-8k\right)}^2+5r^2=816$

$A^2=\left(\begin{array}{ll}\mathrm{c}\mathrm{o}\mathrm{s}?2\theta & i\mathrm{s}\mathrm{i}\mathrm{n}?2\theta \\ i\mathrm{s}\mathrm{i}\mathrm{n}?2\theta & \mathrm{c}\mathrm{o}\mathrm{s}?2\theta \end{array}\right)$

Similarly, $A^5=\left(\begin{array}{cc}\mathrm{c}\mathrm{o}\mathrm{s}?5\theta & i\mathrm{s}\mathrm{i}\mathrm{n}?5\theta \\ i\mathrm{s}\mathrm{i}\mathrm{n}?5\theta & \mathrm{c}\mathrm{o}\mathrm{s}?5\theta \end{array}\right)=\left(\begin{array}{ll}a& b\\ c& d\end{array}\right)$

(1) $a^2+b^2={\mathrm{c}\mathrm{o}\mathrm{s}}^2?5\theta -{\mathrm{s}\mathrm{i}\mathrm{n}}^2?5\theta =\mathrm{c}\mathrm{o}\mathrm{s}?10\theta =\mathrm{c}\mathrm{o}\mathrm{s}?{75}^{?}$

(2) $a^2-d^2={\mathrm{c}\mathrm{o}\mathrm{s}}^2?5\theta -{\mathrm{c}\mathrm{o}\mathrm{s}}^2?5\theta =0$

(3) $a^2-b^2={\mathrm{c}\mathrm{o}\mathrm{s}}^2?5\theta +{\mathrm{s}\mathrm{i}\mathrm{n}}^2?5\theta =1$

(4) $a^2-c^2={\mathrm{c}\mathrm{o}\mathrm{s}}^2?5\theta +{\mathrm{s}\mathrm{i}\mathrm{n}}^2?5\theta =1$

P (E_n) = n/36 for n = 1, 2, 3, …., 8

$P\left(A\right)=\frac{Anypossiblesumof\left(1,2,3,.........,8\right)\left(=\alpha say\right)}{36}$

$\frac{\alpha }{36}\ge \frac{4}{5}$

$\therefore a\ge 29$

If one of the number from {1, 2, ….8} is left then total $\ge$ 29 by 3 ways

Similarly by leaving terms more 2 or 3 we get 16 more combinations

$\therefore$ Total number of different set a possible is 16 + 3

= 19

The required probability

$=\frac{AreaofRegionPQCAP}{AreaofRegionABCA}$

$=\frac{\frac{1}{2}*8*6-\frac{1}{2}*2*4}{\frac{1}{2}*8*6}$

$=\frac{5}{6}$

This is a  Fill in the Blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}?P\left(A/B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}\\ \Rightarrow P\left(A\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}\\ \Rightarrow P\left(A\cap B\right)=P\left(A\right).P\left(B\right)\\ So,AisindependentofB.\end{array}$