Coursework solutions for Math 103 Probability: Week 12

1. Let $A$ denote the event ‘Abel has the live bullet’. Similarly define $B$ and $C$, and let $H$ denote the event ‘the victim is hit’.
   

   [1 for setting up the events correctly]

   Clearly $\mathrm{P}(A)=\mathrm{P}(B)=\mathrm{P}(C)=1/3$.

   1. As $\{A\},\{B\},\{C\}$ partitions the sample space, the law of total probability gives

      $\mathrm{P}(H)$ $=$ $\mathrm{P}(H|A)\mathrm{P}(A)+\mathrm{P}(H|B)\mathrm{P}(B)+\mathrm{P}(H|C)\mathrm{P}(C)$

      $=$ $(0.6+0.7+0.8)/3=0.7.$

      
      

      [0.5 for invoking law of total prob (either writing it in abstract, or stating the name)]

      
      

      [0.5 for putting correct numbers into the law]

   2. By Bayes theorem $\mathrm{P}(C|H)=\mathrm{P}(H|C)\mathrm{P}(C)/\mathrm{P}(H)=\frac{0.8\times 1/3}{0.7}=0.381$.
       
      

      [0.5 for knowing we want $\mathrm{P}(C|H)$, 0.5 for using Bayes’ theorem]

2. $\mathrm{\Omega }=\{BB,BY,BG,YB,YY,YG,GB,GY,GG\}$ and

   $R(BB)$ $=$ $-2,$

   $R(YY)$ $=$ $4,$

   $R(GG)$ $=$ $0,$

   $R(BY)$ $=$ $R(YB)=1,$

   $R(BG)$ $=$ $R(GB)=-1,$

   $R(YG)$ $=$ $R(GY)=2.$

   The induced sample space for $R$ is $𝒮=\{-2,-1,0,1,2,4\}$.
   

   [1 for $\mathrm{\Omega }$, 0.5 for correct specification of $R$, 1 for $𝒮$]

   Now

   $\mathrm{P}(R=-2)$ $=$ $\mathrm{P}(\{BB\})={\displaystyle \frac{8.7}{14.13}}={\displaystyle \frac{4}{13}},$

   $\mathrm{P}(R=-1)$ $=$ $\mathrm{P}(\{BG,GB\})={\displaystyle \frac{8.2}{14.13}}+{\displaystyle \frac{2.8}{14.13}}={\displaystyle \frac{16}{91}},$

   $\mathrm{P}(R=0)$ $=$ $\mathrm{P}(\{GG\})={\displaystyle \frac{2.1}{14.13}}={\displaystyle \frac{1}{91}},$

   $\mathrm{P}(R=1)$ $=$ $\mathrm{P}(\{BY,YB\})={\displaystyle \frac{8.4}{14.13}}+{\displaystyle \frac{4.8}{14.13}}={\displaystyle \frac{32}{91}},$

   $\mathrm{P}(R=2)$ $=$ $\mathrm{P}(\{YG,GY\})={\displaystyle \frac{4.2}{14.13}}+{\displaystyle \frac{2.4}{14.13}}={\displaystyle \frac{8}{91}},$

   $\mathrm{P}(R=4)$ $=$ $\mathrm{P}(\{YY\})={\displaystyle \frac{4.3}{14.13}}={\displaystyle \frac{6}{91}}.$

   [0.25 for each correct probability]

3. The sets $\{R=r\}$, $r=0,1,2,\mathrm{\dots }$ partition $\mathrm{\Omega }$.

   [1]

   Therefore

   ${\displaystyle \sum _{\omega \in \mathrm{\Omega }}}R(\omega )\mathrm{P}(\{\omega \})$ $=$ ${\displaystyle \sum _{r=0}^{\mathrm{\infty }}}{\displaystyle \sum _{\omega \in \{R=r\}}}R(\omega )\mathrm{P}(\{\omega \})$

   $=$ ${\displaystyle \sum _{r=0}^{\mathrm{\infty }}}r{\displaystyle \sum _{\omega \in \{R=r\}}}\mathrm{P}(\{\omega \})$

   $=$ ${\displaystyle \sum _{r=0}^{\mathrm{\infty }}}r\mathrm{P}(R=r)$

   $=$ ${\displaystyle \sum _{r=0}^{\mathrm{\infty }}}r{p}_R(r).$

   [2]