Let $|X|$ denote the number of elements in set $X$. Let $S = \{1, 2, 3, 4, 5, 6\}$ be a sample space,where each element is equally likely to occur. If $A$ and $B$ are independent events associated with $S$,then the number of ordered pairs $(A, B)$ such that $1 \leq |B| < |A|$ equals:

$U_1, U_2, U_3$ are three urns. $U_1$ contains $5$ red,$3$ white,$2$ black balls; $U_2$ contains $4$ red,$4$ white,$2$ black balls and $U_3$ contains $3$ red,$4$ white,$3$ black balls. If a ball is chosen at random from an urn chosen at random,then the probability of not getting a black ball is

If $A, B$ and $C$ are three independent events such that $P(A)=P(B)=P(C)=P$, then $P$ (at least two of $A, B$ and $C$ occur) is equal to

$A$ random variable $X$ has the following probability distribution:
| $X=x$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| $P(X=x)$ | $0.15$ | $0.23$ | $0.12$ | $0.20$ | $0.08$ | $0.10$ | $0.05$ | $0.07$ |
For the events $E = \{X \text{ is a prime number}\}$ and $F = \{X < 5\}$,find $P(E \cup F)$.

Let $X$ and $Y$ be two events such that $P(X \mid Y)=\frac{1}{2}$,$P(Y \mid X)=\frac{1}{3}$,and $P(X \cap Y)=\frac{1}{6}$. Which of the following is (are) correct?
$(A)$ $P(X \cup Y)=\frac{2}{3}$
$(B)$ $X$ and $Y$ are independent
$(C)$ $X$ and $Y$ are not independent
$(D)$ $P(X^C \cap Y)=\frac{1}{3}$

If $a$ and $b$ are chosen randomly from the set $\{1, 2, 3, 4, 5, 6\}$ with replacement,then the probability that $\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{{a^x} + {b^x}}}{2}} \right)^{\frac{2}{x}}}=6$ is