Probability

11. The sample space of given condition is

A $S=\{1,2,3,4,5,6\}$

$E=\{1,3,5\}$ $,F=\{2,3\}$ $,G=\{2,3,4,5\}$

$P(E)=\frac{3}{6}=\frac{1}{2}$ $,P(F)=\frac{2}{6}=\frac{1}{3}$ $P(G)=\frac{4}{6}=\frac{2}{3}$

$\mathrm{(i)\; P}(E/F)$ and $P(F/E)$

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

$P(E/F)=\frac{P(E\cap F)}{P(F)}=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=\frac{1}{2}$

and $P(F/E)=\frac{P(F\cap E)}{P(E)}=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{1}{3}$

$\mathrm{(ii)\; P}(E/G)$ and $P(G/E)$

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

$P(E/G)=\frac{P(E\cap G)}{P(G)}=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=\frac{1}{2}$

$P(G/E)=\frac{P(G\cap E)}{P(E)}=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{2}{3}$

$\mathrm{(iii)\; P}((E\cup F)/G)$ and $P((E\cap F)/G)$

$(E\cup F)=\{1,2,3,5\}$

$(E\cup F)\cap G=\{1,2,3,5\}\cap \{2,3,4,5\}$

$=\{2,3,5\}$

and $E\cap F=\left\{3\right\}$

$(E\cap F)\cap G=\left\{3\right\}$

$P(E\cup G)=\frac{4}{6}=\frac{2}{3}$

Now,

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

So,

$P((E\cup F)/G)=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=\frac{3}{4}$

$P((E\cap F)/G)=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=\frac{3}{12}=\frac{1}{4}$

10. When two die are rolled, the sample space has $6*6=36$

Let $A$ be obtaining a sum greater than

$9=\{(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)\}$

And $B$ be black die result in $5$ ,

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

$A\cap B=\{(5,5),(5,6)\}$

$\mathrm{a.\; P}(A/E)=\frac{P(A\cap B)}{P(B)}=\frac{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$36$}\right.}{\raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$36$}\right.}=\frac{1}{3}$

b. Let,

$E:$ sum of observation is $8$

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

$F:$ Red die resulted in number less than $4$

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

$E\cap F=\{(5,3),(6,2)\}$

$P(F)=\frac{18}{36}$ and $P(E\cap F)=\frac{2}{36}$

$P(E/F)=\frac{P(E\cap F)}{P(F)}=\frac{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$36$}\right.}{\raisebox{1ex}{$18$}\!\left/ \!\raisebox{-1ex}{$36$}\right.}=\frac{2}{18}=\frac{1}{9}$

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,

$P(B)=0.5$

$P(A\cap B)=0.32$

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