Let A and B be two independent events such that $$P(A) = \frac{1}{3}$$ and $$P(B) = \frac{1}{6}$$. Then, which of the following is true?

We are given two independent events A and B with probabilities

$$P(A)=\frac13, \qquad P(B)=\frac16.$$

Because A and B are independent, the fundamental fact we shall use again and again is

$$P(A\cap B)=P(A)\,P(B).$$

Independence also carries over to complements; that is, A is independent of B′ and A′ is independent of both B and B′. Now we test each option one by one.

Option A asks for the conditional probability $$P\!\left(A\;|\;B\right).$$ The definition of conditional probability is

$$P\!\left(A\;|\;B\right)=\frac{P(A\cap B)}{P(B)}.$$

Substituting the values, we have

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

so

$$P\!\left(A\;|\;B\right)=\frac{\dfrac1{18}}{\dfrac16}=\frac1{18}\times\frac61=\frac6{18}=\frac13.$$

The option claims $$\frac23,$$ therefore Option A is false.

Option B requires the conditional probability $$P\!\left(A\;|\;B'\right),$$ where B′ is the complement of B. First, evaluate $$P(B')$$:

$$P(B')=1-P(B)=1-\frac16=\frac56.$$

Because A is independent of B′, we write

$$P(A\cap B')=P(A)\,P(B')=\frac13\cdot\frac56=\frac5{18}.$$

Applying the conditional-probability formula,

$$P\!\left(A\;|\;B'\right)=\frac{P(A\cap B')}{P(B')}=\frac{\dfrac5{18}}{\dfrac56}=\frac5{18}\times\frac65=\frac6{18}=\frac13.$$

This exactly matches the value stated in Option B, so Option B is true.

Option C asks for $$P\!\left(A'\;|\;B'\right).$$ First compute $$P(A')$$:

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

Since A′ and B′ are independent,

$$P(A'\cap B')=P(A')\,P(B')=\frac23\cdot\frac56=\frac{10}{18}=\frac59.$$

Then

$$P\!\left(A'\;|\;B'\right)=\frac{P(A'\cap B')}{P(B')}=\frac{\dfrac59}{\dfrac56}=\frac59\times\frac65=\frac{30}{45}=\frac23.$$

Option C claims $$\frac13,$$ so it is incorrect.

Option D involves $$P\!\left(A\;|\;A\cup B\right).$$ First, notice that

$$A\cap(A\cup B)=A,$$

so the numerator of the conditional probability is simply $$P(A)=\frac13.$$ Next we need $$P(A\cup B).$$ Using the addition rule,

$$P(A\cup B)=P(A)+P(B)-P(A\cap B).$$

We already have $$P(A\cap B)=\frac1{18},$$ hence

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

Now compute the conditional probability:

$$P\!\left(A\;|\;A\cup B\right)=\frac{P(A)}{P(A\cup B)}=\frac{\dfrac13}{\dfrac49}=\frac13\times\frac94=\frac{9}{12}=\frac34.$$

The option lists $$\frac14,$$ so Option D is false.

Only Option B agrees with the actual calculation.

Hence, the correct answer is Option 2.