In a class of $125$ students,$70$ passed in Mathematics,$55$ in Statistics,and $30$ in both. The probability that a student selected at random from the class has passed in only one subject is

Suppose $A_1, A_2, A_3, \dots, A_{30}$ are $30$ sets each having $5$ elements and $B_1, B_2, \dots, B_n$ are $n$ sets each with $3$ elements. Let $\bigcup_{i=1}^{30} A_i = \bigcup_{j=1}^n B_j = S$ and each element of $S$ belongs to exactly $10$ of the $A_i$'s and exactly $9$ of the $B_j$'s. Then $n$ is equal to:

In a class of $100$ students,$15$ students chose only physics (but not mathematics and chemistry),$3$ chose only chemistry (but not mathematics and physics),and $45$ chose only mathematics (but not physics and chemistry). Of the remaining students,it is found that $23$ have taken physics and chemistry,$20$ have taken physics and mathematics,and $12$ have taken mathematics and chemistry. The number of students who chose all the three subjects is

Let $A = \{1, 2, 3, 4, 5, 6, 7\}$. Define $B = \{T \subseteq A : \text{either } 1 \notin T \text{ or } 2 \in T\}$ and $C = \{T \subseteq A : \text{the sum of all the elements of } T \text{ is a prime number}\}$. Then the number of elements in the set $B \cup C$ is $\dots\dots$

In a school,there are $21$ students in the cricket team,$26$ in the hockey team,and $29$ in the football team. Among these,$14$ play both hockey and cricket,$15$ play both hockey and football,and $12$ play both football and cricket. If $8$ students play all three games,find the total number of students in the three athletic teams.

Let $S = \{1, 2, 3, \ldots, 40\}$ and let $A$ be a subset of $S$ such that no two elements in $A$ have their sum divisible by $5$. What is the maximum number of elements possible in $A$?