For two given events $A$ and $B$, $P\,(A \cap B) = $

Let $A$ and $B $ be two events such that  $P\left( {\overline {A \cup B} } \right) = \frac{1}{6}\;,P\left( {A \cap B} \right) = \frac{1}{4}$ and $P\left( {\bar A} \right) = \frac{1}{4}$ where $\bar A$ stands for the complement of the event $A$. Then the events $A$ and$B$ are

In two events $P(A \cup B) = 5/6$, $P({A^c}) = 5/6$, $P(B) = 2/3,$ then $A$ and $B$ are

Two persons $A$ and $B$ throw a (fair)die (six-faced cube with faces numbered from $1$ to $6$ ) alternately, starting with $A$. The first person to get an outcome different from the previous one by the opponent wins. The probability that $B$ wins is

If an integer is chosen at random from first $100$ positive integers, then the probability that the chosen number is a multiple of $4$ or $6$, is

If $A$ and $B$ are events such that $P(A \cup B) = 3/4,$ $P(A \cap B) = 1/4,$ $P(\bar A) = 2/3,$ then $P(\bar A \cap B)$ is