Let X and Y be two events such that $$P(X) = {1 \over 3}$$, $$P(X|Y) = {1 \over 2}$$ and $$P(Y|X) = {2 \over 5}$$. Then

Let $${n_1}$$ and $${n_2}$$ be the number of red and black balls, respectively, in box $${\rm I}$$. Let $${n_3}$$ and $${n_4}$$ be the number of red and black balls, respectively, in box $${\rm I}{\rm I}.$$

One of the two boxes, box $${\rm I}$$ and box $${\rm I}{\rm I},$$ was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability that this red ball was drawn from box $${\rm I}{\rm I}$$ is $${1 \over 3},$$ then the correct option(s) with the possible values of $${n_1}$$ $${n_2},$$ $${n_3}$$ and $${n_4}$$ is (are)

Let $${n_1}$$ and $${n_2}$$ be the number of red and black balls, respectively, in box $${\rm I}$$. Let $${n_3}$$ and $${n_4}$$ be the number of red and black balls, respectively, in box $${\rm I}{\rm I}.$$

A ball is drawn at random from box $${\rm I}$$ and transferred to box $${\rm I}$$$${\rm I}.$$ If the probability of drawing a red ball from box $${\rm I},$$ after this transfer, is $${1 \over 3},$$ then the correct option(s) with the possible values of $${n_1}$$ and $${n_2}$$ is(are)

Let $$X$$ and $$Y$$ be two events such that $$P\left( {X|Y} \right) = {1 \over 2},$$ $$P\left( {Y|X} \right) = {1 \over 3}$$ and $$P\left( {X \cap Y} \right) = {1 \over 6}.$$ Which of the following is (are) correct ?