Each set $X_r$ contains $5$ elements and each set $Y_r$ contains $4$ elements. If $\bigcup_{r=1}^{24} X_r = S = \bigcup_{r=1}^{n} Y_r$,and each element of set $S$ belongs to exactly $10$ of the $X_r$'s and to exactly $6$ of the $Y_r$'s,then find the value of $n$.

Consider a three-digit number $N = 100x + 10y + z$,where $x, y, z$ are the digits at the hundreds,tens,and units places respectively. The number satisfies the following properties:
$I$. If its digits in the units place and tens place are interchanged,the number increases by $36$.
$II$. If its digits in the units place and hundreds place are interchanged,the number decreases by $198$.
If the digits in the tens place and hundreds place are interchanged,then the number:

If $A$ and $B$ are events of a random experiment such that $P(A \cup B) = \frac{3}{4}$,$P(A \cap B) = \frac{1}{4}$,and $P(\bar{A}) = \frac{2}{3}$,then find $P(\bar{A} \cap B)$.

For three non-impossible events $A$,$B$,and $C$,$P(A \cap B \cap C) = 0$,$P(A \cup B \cup C) = \frac{3}{4}$,$P(A \cap B) = \frac{1}{3}$,and $P(C) = \frac{1}{6}$. The probability that exactly one of $A$ or $B$ occurs but $C$ does not occur is:

In a class of $60$ students,$40$ opted for $NCC$,$30$ opted for $NSS$ and $20$ opted for both $NCC$ and $NSS$. If one of these students is selected at random,then the probability that the student selected has opted neither for $NCC$ nor for $NSS$ is

The probability that at least one of $A$ and $B$ occurs is $0.6$. If $A$ and $B$ occur simultaneously with probability $0.3$, then $P(A') + P(B')$ is