1. Interpret the notation

The expressions like $P\left(\frac{A}{B}\right)$ mean conditional probability: $P(A\mid B).$

So the options are:

• A: $P(A\mid A\cup B)=\frac14$
• B: $P(A\mid B)=\frac23$
• C: $P(A'\mid B')=\frac13$
• D: $P(A\mid B')=\frac13$

We are given: $P(A)=\frac13,\qquad P(B)=\frac16,$ and $A,B$ are independent.

Hence, $P(A\cap B)=P(A)P(B)=\frac13\cdot\frac16=\frac1{18}.$

2. Check option A

We need: $P(A\mid A\cup B)=\frac{P(A\cap (A\cup B))}{P(A\cup B)}.$

Since $A\cap (A\cup B)=A$, $P(A\mid A\cup B)=\frac{P(A)}{P(A\cup B)}.$

Now, $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ $=\frac13+\frac16-\frac1{18}$ $=\frac6{18}+\frac3{18}-\frac1{18}=\frac8{18}=\frac49.$

Therefore, $P(A\mid A\cup B)=\frac{1/3}{4/9}=\frac13\cdot\frac94=\frac34.$

So option A is false.

3. Check option B

$P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{1/18}{1/6}=\frac16\cdot\frac61?$ More carefully, $\frac{1/18}{1/6}=\frac{1}{18}\cdot 6=\frac13.$

So option B says $\frac23$, but actual value is $P(A\mid B)=\frac13.$ Hence option B is false.

(Also, since $A$ and $B$ are independent, $P(A\mid B)=P(A)=\frac13$.)

4. Check option C

We need $P(A'\mid B')$.

Because independence is preserved under complements, $A'$ and $B'$ are also independent. Hence, $P(A'\mid B')=P(A').$

Now, $P(A')=1-P(A)=1-\frac13=\frac23.$

So, $P(A'\mid B')=\frac23.$

Option C says $\frac13$, so it is false.

5. Check option D

We need: $P(A\mid B')=\frac{P(A\cap B')}{P(B')}.$

Since $A$ and $B$ are independent, $A$ and $B'$ are also independent. Therefore, $P(A\mid B')=P(A)=\frac13.$

So option D is true.

6. Conclusion

The only true statement is: $\boxed{\text{D}}$