Maths

7. The sample space of given experiment is

$S=\{HH,HT,TH,TT\}$

$\mathrm{(i)\; E}:\{HT,TH\}$

$\Rightarrow F:\{HT,TH\}$

$\Rightarrow E\cap F=\{HT,TH\}$

$P(F)=\frac{2}{4}=\frac{1}{2}$ $;P(E\cap F)=\frac{2}{4}=\frac{1}{2}$

$P(E/F)=\frac{P(E\cap F)}{P(F)}=\frac{\frac{1}{2}}{\frac{1}{2}}=1$

$\mathrm{(ii)\; E}:\left\{HH\right\}$

$F:\left\{TT\right\}$

$E\cap F:\left\{\right\}$

So, $,P(F)=\frac{1}{4}$ $,P(E\cap F)=\frac{0}{4}=0$

$P(E/F)=\frac{P(E\cap F)}{P(F)}=\frac{0}{\frac{1}{4}}=0$

6. The sample space has 8 outcome,

$S=\{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT\}$

$\mathrm{(i)\; E}=\{HHH,HTH,THH,TTH\}$

$F=\{HHH,HHT\}$

and $E\cap F=\left\{HHH\right\}$

$P(F)=\frac{2}{8}$ $;P(E\cap F)=\frac{1}{8}$

$P(E/F)=\frac{P(E\cap F)}{P(F)}=\frac{\frac{1}{8}}{\frac{2}{8}}=\frac{1}{2}$

$\mathrm{(ii)\; E}=\{HHH,HHT,HTH,THH\}$

$F=\{HHT,HTH,HTT,THH,THT,TTH,TTT\}$

$E\cap F=\{HHH,HTH,THH\}$

$P(F)=\frac{7}{8}$ $;P(E\cap F)=\frac{3}{8}$

$P(E/F)=\frac{P(E\cap F)}{P(F)}=\frac{\frac{3}{8}}{\frac{7}{8}}=\frac{3}{7}$

$\mathrm{(iii)\; E}:\{HHH,HHT,HTT,HTH,THH,THT,TTH\}$

$F:\{HHT,HTH,HTT,THH,THT,TTT,TTH\}$

$E\cap F:\{HHT,HTH,HTT,THH,THT,TTH\}$

$P(F)=\frac{7}{8}$ $;P(E\cap F)=\frac{6}{8}$

$P(E/F)=\frac{P(E\cap F)}{P(F)}=\frac{\frac{5}{8}}{\frac{6}{8}}=\frac{6}{7}$

5. Given,

$P(A)=\frac{6}{11}$

$P(B)=\frac{5}{11}$

$P(A\cup B)=\frac{7}{11}$

$\mathrm{(i)\; P}(A\cap B)$

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

$\Rightarrow \frac{7}{11}=\frac{6}{11}+\frac{5}{11}-P(A\cap B)$

$\Rightarrow \frac{7}{11}-\frac{11}{11}=-P(A\cap B)$

$\Rightarrow -\frac{4}{11}=-P(A\cap B)$

$\Rightarrow P(A\cap B)=\frac{4}{11}$

$\mathrm{(ii)\; P}(A/B)$

$P(A/B)=\frac{P(A\cap B)}{P(B)}$

$=\frac{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$11$}\right.}{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$11$}\right.}=\frac{4}{5}$

$\mathrm{(iii)\; P}(B/A)$

$P(B/A)=\frac{P(B\cap A)}{P(A)}$

$=\frac{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$11$}\right.}{\raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$11$}\right.}=\frac{4}{6}=\frac{2}{3}$

4. Given,

$2P(A)=P(B)=\frac{5}{13}$

$P(A/B)=\frac{2}{5}$

$2P(A)=\frac{5}{13}\Rightarrow P(A)=\frac{5}{13*2}=\frac{5}{26}$

$\Rightarrow P(A/B)=\frac{P(A\cap B)}{P(B)}$

$\Rightarrow \frac{2}{5}=\frac{P(A\cap B)}{\frac{5}{13}}$

$\Rightarrow \frac{2}{5}*\frac{5}{13}=P(A\cap B)\Rightarrow P(A\cap B)=\frac{2}{13}$

$\therefore P(A\cup B)=P(A)+P(B)-P(A\cap B)$

$=\frac{5}{26}+\frac{5}{13}-\frac{2}{13}$

$=\frac{5+10-4}{26}=\frac{11}{26}$

3. Given,

$P(A)=0.8$

$P(B)=0.5$

$P(B/A)=0.4$

$\mathrm{(i)\; P}(A\cap B)$

$P(B/A)=\frac{P(B\cap A)}{P(A)}$ $[?P(A\cap B)=P(B\cap A)]$

$\Rightarrow 0.4=\frac{P(A\cap B)}{0.8}$

$\Rightarrow P(A\cap B)=0.4*0.8$

$=0.32$

$\mathrm{(ii)\; P}(A/B)$

$P(A/B)=\frac{P(A\cap B)}{P(B)}$

$P(A/B)=\frac{0.32}{0.5}$

$=0.64$

$\mathrm{(iii)\; P}(A\cup B)$

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

$=0.8+0.5-0.32$

$=0.98$

2. Given,

$P(B)=0.5$

$P(A\cap B)=0.32$

$P(A/B)=\frac{P(A\cap B)}{P(B)}=\frac{0.32}{0.5}=\frac{32}{50}=\frac{16}{45}$

1. Given,

$P(E)=0.6$

$P(F)=0.3$

$P(E\cap F)=0.2$

$P(E/F)=\frac{P(E\cap F)}{P(F)}=\frac{0.2}{0.3}=\frac{2}{3}$

$P(F/E)=\frac{P(F\cap E)}{P(E)}=\frac{0.2}{0.6}=\frac{2}{6}=\frac{1}{3}$

36. For the given inequality, x> –3.

The equation of the line is x= –3.

This line divides the xy-plane into planer I and II. We take a point (0,0) to check the correctness of the inequality.

So, 0> –3 which is true.

So, the solution of the region is I which includes the origin.

The dotted line indicates that any point on the line does not satisfy the inequality.

35. For the given equation inequality y< 2,  the equation of line is y= 2

This line devides the xy-plane into two planer I and II. We take a point (0,0) to check the correctness of the inequality.

So, 0< 2

0< 2 which is false.

So, the solution of the region is II which does not include the origin.

The dotted line indicates that any point on the line does not satisfy the inequality.

34. For the inequality 3y – 5x<30, the equation of line is 3y 5x=30. We consider the table below to plot 3y 5x =30.

$\begin{array}{l}x\\ y\end{array}|\begin{array}{l}0\\ 10\end{array}|\begin{array}{l}-6\\ 0\end{array}|$

This line divides the xy-plane into two planer I and II. We select a point (0,0) and check the correctness the inequality.

3 * 0 – 5 * 0 < 30

0 < 30 which is true.

So, the solution region is I which includes the origin. The dotted line indicates that any point on the line will not satisfy the given inequality.