Quiz solutions for Math 103 Probability: Week 12

1. Find $c$ (D)

   $1={\displaystyle \sum _{r=0}^{10}}cr=c{\displaystyle \frac{11\times 10}{2}}=55c.$

2. Dice (D)
   $\mathrm{\Omega }=\{a_1{a}_2:a_i=1,2,\mathrm{\dots },6\}$.

   $\mathrm{P}(R\text{ even})$ $=$ $\mathrm{P}(\{R=2\}\cup \{R=4\}\cup \{R=6\})$

   $=$ $\mathrm{P}(R=2)+\mathrm{P}(R=4)+\mathrm{P}(R=6)\mathit{\hspace{1em}}\text{by Axiom 3}$

   $=$ $\mathrm{P}(\{12,21,22\})+\mathrm{P}(\{14,41,24,42,34,43,44\})+\mathrm{P}(\{16,61,26,62,36,63,46,64,56,65,66\})$

   $=$ ${\displaystyle \frac{21}{36}}={\displaystyle \frac{7}{12}}.$

3. Variance (B)
   $\mathrm{Var}(R)=\mathrm{E}[R^2]-{(\mathrm{E}[R])}^2$, so we need to calculate the two terms on the right hand side.
   $\mathrm{E}[R]=(1+2+\mathrm{\dots }+10)\times \frac{1}{10}=5.5$ and $\mathrm{E}[R^2]=(1^2+2^2+\mathrm{\dots }+{10}^2)\times \frac{1}{10}=38.5$.
   Therefore $\mathrm{Var}(R)=38.5-{(5.5)}^2=8.25.$

4. Seeded Dice Throws (E)
   The code needed is

   set.seed(565)

   rolls <-sample(1:6,200,replace=TRUE,prob=rep(1/6,6))

   sum(rolls==5)

   This gives answer 35.

5. Simulating the Lottery (C)
   The sample space needs to be 1,2,..,59. The size needs to be 6.

   We sample without replacement i.e. replace=FALSE, since balls are not put back into the lottery once sampled.

   The probability set is (1/59,…,1/59) since all balls are equally likely to be drawn.

   Hence the answer is (C)