Coursework solutions for Math 103 Probability: Week 11

1. 2 marks Let A = ‘Andy wins’ and B = ‘Barry wins’. The outcomes of the match can be represented as a path through the tree:
   
   The elementary outcomes are therefore the letters passed between the start and a leaf of the tree:
   $\mathrm{\Omega }=\{AA,ABAA,ABABA,ABABB,ABB,BB,BABB,BABAB,BABAA,BAA\}$.

   [1 mark for understanding the logic and translating into something similar to the tree. 1 mark for clear description of the sample space.]

2. 4 marks

   Since $A\subseteq B$, $B=A\cup (B\cap A^c)$.
   

   [1 mark for this decomposition]

   As $A$ and $B\cap A^c$ are disjoint, Axiom 3 tells us that

   $$\mathrm{P}(B)=\mathrm{P}(A)+\mathrm{P}(B\cap A^c)$$

   [0.5 mark for invoking Axiom 3, 0.5 mark for the conclusion]

   By Axiom 1, $\mathrm{P}(B\cap A^c)\ge 0$.
   

   [0.5 mark for invoking Axiom 1, 0.5 mark for the conclusion]

   Therefore

   $$\mathrm{P}(B)=\mathrm{P}(A)+\mathrm{P}(B\cap A^c)\ge \mathrm{P}(A).$$

   [1 mark for expressing the conclusion correctly and neatly]

3. 4 marks

   1. $\mathrm{P}(B|A^c)$ is the probability of developing breast cancer (event $B$) if we know that the mammogram was not positive (event $A^c$). From the text we know that this is 20 in 100,000. Therefore

      $$\mathrm{P}(B|A^c)=\frac{20}{100,000}=0.0002.$$

      [1 mark]

      Similarly, $\mathrm{P}(B|A)$ is the probability of developing breast cancer (event $B$) if we know that the mammogram was positive (event $A$). From the text we know this was 1 in 10. Therefore

      $$\mathrm{P}(B|A)=\frac{1}{10}=0.1.$$

      [1 mark]

   2. By the law of total probability from notes,

      $$\mathrm{P}(B)=\mathrm{P}(B|A)P(A)+\mathrm{P}(B|A^c)P(A^c)$$

      [0.5 marks, not awarded if jump straight to numbers without justification]

      We know $\mathrm{P}(B|A)$ and $\mathrm{P}(B|A^c)$. We also know from the text that $\mathrm{P}(A)=0.07$, so $\mathrm{P}(A^c)=1-\mathrm{P}(A)=0.93$.
      

      [0.5 marks each for $\mathrm{P}(A)$ and $\mathrm{P}(A^c)$]

      Hence

      $\mathrm{P}(B)$ $=$ $0.1\times .07+.0002\times .93$

      $=$ $.007186.$

      [0.5 mark for putting the numbers in (even if the previously calculated numbers were incorrect)]