Step-by-step Solution:

We are given the following probabilities for two events X and Y:

1. $P\left( {X|Y} \right) = {1 \over 2}$
2. $P\left( {Y|X} \right) = {1 \over 3}$
3. $P\left( {X \cap Y} \right) = {1 \over 6}$

We will evaluate each option based on these given values.

Step 1: Find P(X) and P(Y)

Using the definition of conditional probability, $P\left( {A|B} \right) = {{P\left( {A \cap B} \right)} \over {P\left( B \right)}}$.

For $P\left( {X|Y} \right)$: $P\left( {X|Y} \right) = {{P\left( {X \cap Y} \right)} \over {P\left( Y \right)}}$ ${1 \over 2} = {{{1 \over 6}} \over {P\left( Y \right)}}$ $P\left( Y \right) = {{{1 \over 6}} \over {{1 \over 2}}} = {1 \over 6} \times 2 = {2 \over 6} = {1 \over 3}$

For $P\left( {Y|X} \right)$: $P\left( {Y|X} \right) = {{P\left( {Y \cap X} \right)} \over {P\left( X \right)}}$ ${1 \over 3} = {{{1 \over 6}} \over {P\left( X \right)}}$ $P\left( X \right) = {{{1 \over 6}} \over {{1 \over 3}}} = {1 \over 6} \times 3 = {3 \over 6} = {1 \over 2}$

So, we have $P\left( X \right) = {1 \over 2}$ and $P\left( Y \right) = {1 \over 3}$.

Step 2: Evaluate Option A

The option is $P\left( {X \cup Y} \right) = {2 \over 3}$. Using the addition rule for probability: $P\left( {X \cup Y} \right) = P\left( X \right) + P\left( Y \right) - P\left( {X \cap Y} \right)$ Substituting the values we found and were given: $P\left( {X \cup Y} \right) = {1 \over 2} + {1 \over 3} - {1 \over 6}$ To add/subtract the fractions, we find a common denominator, which is 6: $P\left( {X \cup Y} \right) = {3 \over 6} + {2 \over 6} - {1 \over 6} = {{3 + 2 - 1} \over 6} = {4 \over 6} = {2 \over 3}$ Therefore, Option A is correct.

Step 3: Evaluate Options B and C

Option B states that X and Y are independent. Option C states that X and Y are not independent.

Two events X and Y are independent if and only if $P\left( {X \cap Y} \right) = P\left( X \right)P\left( Y \right)$.

Let's calculate $P\left( X \right)P\left( Y \right)$: $P\left( X \right)P\left( Y \right) = \left( {{1 \over 2}} \right) \times \left( {{1 \over 3}} \right) = {1 \over 6}$ We are given that $P\left( {X \cap Y} \right) = {1 \over 6}$. Since $P\left( {X \cap Y} \right) = P\left( X \right)P\left( Y \right)$, the events X and Y are independent.

Therefore, Option B is correct and Option C is incorrect.

Alternatively, independence can be checked using conditional probabilities. Events are independent if $P(X|Y) = P(X)$ and $P(Y|X) = P(Y)$. We are given $P(X|Y) = 1/2$, and we calculated $P(X) = 1/2$. They are equal. We are given $P(Y|X) = 1/3$, and we calculated $P(Y) = 1/3$. They are equal. This confirms that X and Y are independent.

Step 4: Evaluate Option D

The option is $P\left( {{X^c} \cap Y} \right) = {1 \over 3}$. $X^c$ denotes the complement of event X. The event $X^c \cap Y$ means that Y occurs but X does not. The formula for this probability is: $P\left( {{X^c} \cap Y} \right) = P\left( Y \right) - P\left( {X \cap Y} \right)$ Substituting the known values: $P\left( {{X^c} \cap Y} \right) = {1 \over 3} - {1 \over 6} = {2 \over 6} - {1 \over 6} = {1 \over 6}$ Since we found $P\left( {{X^c} \cap Y} \right) = {1 \over 6}$, the statement in Option D is incorrect.

Alternatively, since X and Y are independent, $X^c$ and Y are also independent. Therefore, $P(X^c \cap Y) = P(X^c)P(Y)$. $P(X^c) = 1 - P(X) = 1 - 1/2 = 1/2$. $P(X^c \cap Y) = (1/2) \times (1/3) = 1/6$. This again shows that Option D is incorrect.

Conclusion:

Based on the analysis, the correct options are A and B.