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

An event has odds in favour $4 : 5$, then the probability that event occurs, 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 $P\,({A_1} \cup {A_2}) = 1 - P(A_1^c)\,P(A_2^c)$ where $c$ stands for complement, then the events ${A_1}$ and ${A_2}$ are

If $A$ and $B$ an two events such that $P\,(A \cup B) = \frac{5}{6}$,$P\,(A \cap B) = \frac{1}{3}$ and $P\,(\bar B) = \frac{1}{3},$ then $P\,(A) = $

Consider three sets $E_1=\{1,2,3\}, F_1=\{1,3,4\}$ and $G_1=\{2,3,4,5\}$. Two elements are chosen at random, without replacement, from the set $E _1$, and let $S _1$ denote the set of these chosen elements.

Let $E_2=E_1-S_1$ and $F_2=F_1 \cup S_1$. Now two elements are chosen at random, without replacement, from the set $F_2$ and let $S_2$ denote the set of these chosen elements.

Let $G _2= G _1 \cup S _2$. Finally, two elements are chosen at random, without replacement, from the set $G _2$ and let $S _3$ denote the set of these chosen elements.

Let $E_3=E_2 \cup S_3$. Given that $E_1=E_3$, let $p$ be the conditional probability of the event $S_1=\{1,2\}$. Then the value of $p$ is