An event has odds in favour $4 : 5$, then the probability that event occurs, is

A coin is tossed twice. If events $A$ and $B$ are defined as :$A =$ head on first toss, $B = $ head on second toss. Then the probability of $A \cup B = $

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

If $A$ and $B$ are two independent events such that $P\,(A) = 0.40,\,\,P\,(B) = 0.50.$ Find $P$ (neither $A$ nor $B$)

The probabilities of three events $A , B$ and $C$ are given by $P ( A )=0.6, P ( B )=0.4$ and $P ( C )=0.5$ If $P ( A \cup B )=0.8, P ( A \cap C )=0.3, P ( A \cap B \cap$ $C)=0.2, P(B \cap C)=\beta$ and $P(A \cup B \cup C)=\alpha$ where $0.85 \leq \alpha \leq 0.95,$ then $\beta$ lies in the interval

Let $E$ and $F$ be two independent events. The probability that both $E$ and $F$ happens is $\frac{1}{{12}}$ and the probability that neither $E$ nor $F$ happens is $\frac{1}{2},$ then