If $P(A) = \frac{2}{5}$,$P(B) = \frac{1}{4}$ and $P(A \cup B) = \frac{1}{2}$,then $P(A' \cup 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:

Let $S = \{4, 6, 9\}$ and $T = \{9, 10, 11, \ldots, 1000\}$. If $A = \{a_{1} + a_{2} + \ldots + a_{k} : k \in N, a_{1}, a_{2}, \ldots, a_{k} \in S\}$,then the sum of all the elements in the set $T - A$ is equal to:

Let $A = \{x : |x^{2} - 10| \le 6\}$ and $B = \{x : |x - 2| > 1\}$. Then:

$A$ number is selected at random from the set $\{1, 2, 3, 4, \ldots, 1000\}$. What is the probability of getting a number which is a perfect cube or a natural number having an odd number of divisors?

$A$ survey shows that $63 \%$ of the people in a city read newspaper $A$ whereas $76 \%$ read newspaper $B$. If $x \%$ of the people read both the newspapers,then a possible value of $x$ can be