The probability that at least one of the events $E_1$ and $E_2$ occurs is $0.6$. If the probability of the simultaneous occurrence of $E_1$ and $E_2$ is $0.2$,then $P(E_1') + P(E_2') = $

The set $\{x \in R : [x - |x|] = 5\}$ is equal to

Let $S = \{p_1, p_2, \ldots, p_{10}\}$ be the set of the first ten prime numbers. Let $A = S \cup P$,where $P$ is the set of all possible products of distinct elements of $S$. Then the number of all ordered pairs $(x, y)$,where $x \in S$ and $y \in A$,such that $x$ divides $y$,is . . . . . .

If $n(U) = 600$,$n(A) = 100$,$n(B) = 200$,and $n(A \cap B) = 50$,then $n(\bar{A} \cap \bar{B})$ is: ($U$ is the universal set and $A$ and $B$ are subsets of $U$)

Let $S$ be the set of the first $11$ natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \ge 2 \text{ and the product of all elements of } B \text{ is even}\}$ is . . . . . . .

The number of elements in the set $\{n \in \{1, 2, 3, \ldots, 100\} \mid (11)^{n} > (10)^{n} + (9)^{n}\}$ is $.....$