_Examples

L2 p.21

$$\begin{array}{rl}& P(A)=0.28\\ & P(B)=0.06\\ & P(A\cap B)=0.05\end{array}$$ $$\begin{array}{rl}P(A^c\cap B^c)& =P((A\cup B{)}^c)\\ & =1-0.05\\ & =0.95\end{array}$$

L2 p.25

$$\begin{array}{rl}& \\ P(E)=\frac{13\times 3+12}{\left(\genfrac{}{}{0ex}{}{52}{2}\right)}& =\frac{1}{36}\end{array}$$

L2 p.26

$$\begin{array}{rl}P(A)& =\frac{\left(\genfrac{}{}{0ex}{}{4}{2}\right)}{\left(\genfrac{}{}{0ex}{}{52}{2}\right)}\\ P(B)& =\end{array}$$

L2 p.28

$$\begin{array}{rl}& P(A)=\frac{3}{6}=\frac{1}{2}\\ & P(B)=\frac{5}{6}\\ & P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{\frac{2}{6}}{\frac{5}{6}}=\frac{2}{5}\end{array}$$

$P(A\cap B)=\frac{2}{6}$ by counting outcomes, {1, 3, 5} and {2, 3, 4, 5, 6} = {3, 5}

OR using the definition of sample space $B$, size of B is 5 and the no. of outcome of A in B is 2

Then, $P(A\mid B)=\frac{2}{5}$

L2 p.29

A: apple on 1st
B: apple on 2nd

$$\begin{array}{rl}P(A\cap B)& =P(B\mid A)P(A)\\ & =\frac{4}{10}\times \frac{2}{7}\\ & =\frac{4}{35}\end{array}$$

L2 p.31

$$P(E)=\frac{5}{20}\times \frac{4}{19}\times \frac{3}{18}$$