Workshop solutions for Math 103 Probability: Week 11

1. 1. The sample space is the set of all possible outcomes, $\mathrm{\Omega }=\{{\omega }_1,\mathrm{\dots },{\omega }_N\}$.

   2. A sample point is a particular outcome $\omega \in \mathrm{\Omega }$.

   3. An event $A$ is a subset of the possible outcomes contained in the sample space $\mathrm{\Omega }$.

   4. Two events $A$ and $B$ are mutually exclusive if $A\cap B=\mathrm{\varnothing }$.

   5. Sets $A_1,A_2,\mathrm{\dots },A_k$ form a partition of a set $B$ if $A_i\cap A_j=\mathrm{\varnothing }$ for all $i\ne j$ and $B=A_1\cup A_2\cup \mathrm{\dots }\cup A_k$.

2. 	1	2	3	4	5	6
   1	*	*	*	*	*	75
   2	*	*	*	*	7	*
   3	*	*	*	7	*	*
   4	*	*	7	*	*	*
   5	*	7	*	*	*	*
   6	75	*	*	*	*	*

   ‘7’ indicates ‘sum is 7’; ‘5’ indicates ‘difference is 5’.

3. 1. $A\cup (B\cap C)$:

   2. $(A\cup B)\cap (A\cup C)$:

   3. ${(A\cup B)}^c$:

   4. $A^c\cap B^c$:

4. $|\mathrm{\Omega }|=52\times 51=2652$. If $B$ is the event corresponding to a blackjack, then $|B|=4\times 16+16\times 4=128$. So

   $\mathrm{P}(B)={\displaystyle \frac{|B|}{|\mathrm{\Omega }|}}={\displaystyle \frac{128}{2652}}=0.048.$

5. Suppose there are $n$ mice. Then $|\mathrm{\Omega }|=\left(\genfrac{}{}{0pt}{}{n}{4}\right)$. Let $A$ be the event that both mice are chosen and $B$ be the event that neither is chosen. Then $|A|=\left(\genfrac{}{}{0pt}{}{n-2}{2}\right)$ and $|B|=\left(\genfrac{}{}{0pt}{}{n-2}{4}\right)$.

   $\mathrm{P}(A)$ $=$ $2\mathrm{P}(B)$

   $\left({\displaystyle \genfrac{}{}{0pt}{}{n-2}{2}}\right)/\left({\displaystyle \genfrac{}{}{0pt}{}{n}{4}}\right)$ $=$ $2\left({\displaystyle \genfrac{}{}{0pt}{}{n-2}{4}}\right)/\left({\displaystyle \genfrac{}{}{0pt}{}{n}{4}}\right)$

   ${\displaystyle \frac{(n-2)!}{2!(n-4)!}}$ $=$ $2!{\displaystyle \frac{(n-2)!}{4!(n-6)!}}$

   $6$ $=$ $(n-4)(n-5)$

   $n^2-9n+14$ $=$ $0,$

   So $n=2$ or $n=7$. But there are at least 6 mice since 2 white and at least 4 others. Therefore $n=7$.