Workshop solutions for Math 103 Probability: Week 11

1. 1. The three axioms of probability are:

      Axiom 1 (positivity)

      $\mathrm{P}(A)\ge 0$ for all $A\subset \mathrm{\Omega }$.

      Axiom 2 (finitivity)

      $\mathrm{P}(\mathrm{\Omega })=1$.

      Axiom 3 (additivity)

      $\mathrm{P}(A\cup B)=\mathrm{P}(A)+\mathrm{P}(B)$ if $A\cap B=\mathrm{\varnothing }$.

   2. If $A$ and $B$ are two events then, provided $P(B)>0$, the conditional probability of $A$ given $B$ is given by

      $\mathrm{P}(A|B)={\displaystyle \frac{\mathrm{P}(A\cap B)}{\mathrm{P}(B)}}.$

   3. The law of total probability states that

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

   4. Bayes’ theorem states that if $A$ and $B$ are events in the sample space with $\mathrm{P}(A),\mathrm{P}(B)>0$ then

      $\mathrm{P}(B|A)={\displaystyle \frac{\mathrm{P}(A|B)\mathrm{P}(B)}{\mathrm{P}(A)}}.$

   5. $A$ and $B$ are independent events if $\mathrm{P}(A\cap B)=\mathrm{P}(A)\mathrm{P}(B)$.

2. • Axiom 1 is sastisfied since $n_A\ge 0$ and $n>0$.

   • Axiom 2 is satisfied since $P(\mathrm{\Omega })=n/n=1$.

   • Axiom 3 is satisfied since if $A\cap B=\mathrm{\varnothing }$, then $n_{A\cup B}=n_A+n_B$ and so

     $\mathrm{P}(A\cup B)={\displaystyle \frac{n_A+n_B}{n}}={\displaystyle \frac{n_A}{n}}+{\displaystyle \frac{n_B}{n}}=\mathrm{P}(A)+\mathrm{P}(B).$

3. 1. There are $6\times 6=36$ possible outcomes from a single roll of two dice, so $|\mathrm{\Omega }|={36}^n$. If $A$ is the event that a double 6 appears at least once, then by the law of complementary events

      $\mathrm{P}(A)=1-\mathrm{P}(A^c)=1-{\displaystyle \frac{|A^c|}{|\mathrm{\Omega }|}}=1-{\displaystyle \frac{{35}^n}{{36}^n}}.$

   2. $\mathrm{P}(A)>1/2$ if and only if

      So $n>\log 2/(\log (36/35))=24.6$ and hence $n\ge 25$.

4. By the addition law $\mathrm{P}(A)+\mathrm{P}(B)-\mathrm{P}(A\cap B)=\mathrm{P}(A\cup B)\le 1$. So

   $\mathrm{P}(A\cap B)\ge \mathrm{P}(A)+\mathrm{P}(B)-1={\displaystyle \frac{3}{4}}+{\displaystyle \frac{1}{3}}-1={\displaystyle \frac{1}{12}}.$

   Let $\mathrm{\Omega }=\{1,2,\mathrm{\dots },12\}$ with each sample point having equal probability. Let $A=\{1,2,\mathrm{\dots },9\}$ and $B=\{9,10,11,12\}$. Then $\mathrm{P}(A)=3/4$, $\mathrm{P}(B)=1/3$ and $\mathrm{P}(A\cap B)=\mathrm{P}(\{9\})=1/12$.

5. Let $R$ be the event that it rains, $S$ be the event that it snows and $L$ be the event that I am late.

   $\mathrm{P}(R)$ $=$ ${\displaystyle \frac{2}{5}},\mathrm{P}(S)={\displaystyle \frac{3}{5}},$

   $\mathrm{P}(L|R)$ $=$ ${\displaystyle \frac{1}{5}},\mathrm{P}(L|S)={\displaystyle \frac{3}{5}}.$

   1. $\mathrm{P}(L)$ $=$ $\mathrm{P}(L|R)\mathrm{P}(R)+\mathrm{P}(L|S)\mathrm{P}(S)$

      $=$ ${\displaystyle \frac{1}{5}}\times {\displaystyle \frac{2}{5}}+{\displaystyle \frac{3}{5}}\times {\displaystyle \frac{3}{5}}$

      $=$ ${\displaystyle \frac{11}{25}}.$

   2. $$P(S|L)=\frac{P(S\cap L)}{P(L)}=\frac{P(L|S)P(S)}{P(L)}=\frac{\frac{3}{5}\times \frac{3}{5}}{\frac{11}{25}}=\frac{9}{11}.$$