$A$ bag consists of $3$ red balls,$5$ blue balls and $8$ green balls. $A$ ball is selected at random. What is the probability of not getting a blue ball?

Given $P(A \cup B)=0.6$ and $P(A \cap B)=0.2$,the probability that exactly one of the events occurs is:

The probability for a contractor to get a road contract is $\frac{2}{9}$ and to get a building contract is $\frac{5}{9}$. If the probability to get both the contracts is $\frac{1}{6}$,then what is the probability to get neither of these two contracts?

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

Thirty-two persons $X_1, X_2, \ldots, X_{32}$ are randomly seated around a circular table at equal intervals. Two persons $X_i$ and $X_j$ are said to be within earshot of each other if there are at most three persons between them on the minor arc joining $X_i$ and $X_j$. The probability that $X_1$ and $X_3$ are within earshot of each other is:

From a well-shuffled pack of cards,one card is drawn at random. The probability that the card drawn is an ace is