Probability

9. The sample space of given condition is

$S=\{(MFS,MSF,FMS,FSM,SMF,SFM)\}$

$E:\{MFS,FMS,SMF,SFM\}$

$F:\{MFS,SFM\}$

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

$P(E/F)=\frac{P(E\cap F)}{P(F)}=\frac{\frac{1}{3}}{\frac{1}{3}}=1$

If a die is thrown three times, then the number of elements in the sample space will be $6*6*6=216$

Here, $E=$ $\left\{\begin{array}{l}(1,1,4),(1,2,4),(1,3,4),(1,4,4),(1,5,4),(1,6,4)\\ (2,1,4),(2,2,4),(2,3,4),(2,4,4),(2,5,4),(2,6,4)\\ (3,1,4),(3,2,4),(3,3,4),(3,4,4),(3,5,4),(3,6,4)\\ (4,1,4),(4,2,4),(4,3,4),(4,4,4),(4,5,4),(4,6,4)\\ (5,1,4),(5,2,4),(5,3,4),(5,4,4),(5,5,4),(5,6,4)\\ (6,1,4),(6,2,4),(6,3,4),(6,4,4),(6,5,4),(6,6,4)\end{array}\right\}$

$F=\{(6,5,1),(6,5,2),(6,5,3),(6,5,4),(6,5,5),(6,5,6)\}$

$?E?F=\{(6,5,4)\}$

$i.e.,P(F)=\frac{6}{216}$ $,P(E?F)=\frac{1}{216}$

$P(E/F)=\frac{P(E?F)}{P(F)}=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$216$}\right.}{\raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$216$}\right.}=\frac{1}{6}$

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,

$P(E)=0.6$

$P(F)=0.3$

$P(E\cap F)=0.2$

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

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

54. Since, the lock can be open by a combination of four digits from the given ten digits I e, from 0 to 9. The number of ways of selecting 4 digits, = 10C₄

This combination of 4 digits can again be arranged within themselves in 41 ways. So, total number of