The sample space $S$is the whole numbers starting at one and less than $20$.

a.$S=\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$

Let event $A=$the even numbers and event $B=$numbers greater than 13.

b.$A=\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_,B=\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$

c.$P\left(A\right)=\_\_\_\_\_\_\_\_\_\_\_\_\_,P\left(B\right)=\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$

d.$AANDB=\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_,AORB=\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$

e.$P(AANDB)=\_\_\_\_\_\_\_\_\_,P(AORB)=\_\_\_\_\_\_\_\_\_\_\_\_\_$

f.$A\prime =\_\_\_\_\_\_\_\_\_\_\_\_\_,P(A\prime )=\_\_\_\_\_\_\_\_\_\_\_\_\_$

g.$P\left(A\right)+P(A\prime )=\_\_\_\_\_\_\_\_\_\_\_\_$

h. $P\left(A\right|B)=\_\_\_\_\_\_\_\_\_\_\_,P(B\left|A\right)=\_\_\_\_\_\_\_\_\_\_\_\_\_;$are the probabilities equal?

The whole numbers starting at $1$and less than $20$is given by:

$1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19$

$A=\{2,4,6,8,10,12,14,16,18\}$

$B=\{14,15,16,17,18,19\}$

Sample space $S$is given by the whole numbers starting from one and less than $20$:

$S=\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19\}$

As per the given information from the sample space $S$, event $AandB$are as follows:

$$\begin{gathered} A=\{2,4,6,8,10,12,14,16,18\} \\ B=\{14,15,16,17,18,19\} \end{gathered}$$

Probability of $A$is given by:

$P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}$

From the solutions of part a and part b, $n\left(S\right)=19,n\left(A\right)=9andn\left(B\right)=6$.

$\therefore P\left(A\right)=\frac{9}{19}$

Probability of B is given by:

$P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}=\frac{6}{19}$

$$\begin{gathered} AANDB=A\cap B \\ AANDB=\{2,4,6,8,10,12,14,16,18\}\cap \{14,15,16,17,18,19\} \\ AANDB=\{14,16,18\} \end{gathered}$$

$$\begin{gathered} AORB=A\cup B \\ AORB=\{2,4,6,8,10,12,14,16,18\}\cup \{14,15,16,17,18,19\} \\ AORB=\{2,4,6,8,10,12,14,15,16,17,18,19\} \end{gathered}$$

From the solutions of part d, we get $n(AANDB)=3andn(AORB)=12.$

The probability of $AANDB$is given by

$P(AANDB)=\frac{n(AANDB)}{n\left(S\right)}=\frac{3}{19}$

The probability of $AORB$is given by

$P(AORB)=\frac{n(AORB)}{n\left(S\right)}=\frac{12}{19}$

$$\begin{gathered} A^{\text{'}}=S-A \\ {A}^{\text{'}}=\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19\}-\{2,4,6,8,10,12,14,16,18\} \\ {A}^{\text{'}}=\{1,3,5,7,9,11,13,15,17,19\} \end{gathered}$$

From the above solution, $n\left(A^{\text{'}}\right)=10$

$P(A\text{'})=\frac{n(A\text{'})}{n\left(S\right)}=\frac{10}{19}$

From the solutions of part c and part f, we get

$P\left(A\right)+P(A\text{'})=\frac{9}{19}+\frac{10}{19}=\frac{19}{19}=1$

From the solutions of part e and part b, we get

$P(A\mid B)=\frac{P\left(A\mathrm{AND}B\right)}{P\left(B\right)}=\frac{3}{6}$

$P(B\mid A)=\frac{P\left(A\mathrm{AND}B\right)}{P\left(A\right)}=\frac{3}{9}$

Therefore the probabilities are not equal.