Two friends $A$ and $B$ meet every weekend either at a party or at a Sports Club. The probability that they meet at the Sports Club is $\frac{4}{9}$. The probability that they dine together at a party and at the Club are respectively $\frac{1}{3}$ and $\frac{2}{5}$. On a certain weekend, the probability that they disperse without dining together is:

Consider the following statements:
Assertion $(A)$: If $P_1, P_2, P_3$ are probabilities of occurrence of three independent events, then the probability of occurrence of at least one of them is $1 - [(1 - P_1)(1 - P_2)(1 - P_3)]$.
Reason $(R)$: For any three independent events $A, B$, and $C$, $P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A)P(B) - P(A)P(C) - P(B)P(C) + P(A)P(B)P(C)$.
The correct option among the following is:

First bag contains $3$ red and $5$ black balls and second bag contains $6$ red and $4$ black balls. $A$ ball is drawn from each bag. The probability that one ball is red and the other is black,is

$A$ and $B$ alternately throw a pair of dice. $A$ wins if he throws a sum of $5$ before $B$ throws a sum of $8$,and $B$ wins if he throws a sum of $8$ before $A$ throws a sum of $5$. The probability that $A$ wins,if $A$ makes the first throw,is

Let $E$ and $F$ be two independent events. The probability that exactly one of them occurs is $\frac{11}{25}$ and the probability of none of them occurring is $\frac{2}{25}$. If $P(T)$ denotes the probability of occurrence of the event $T$,then which of the following is true?
$(A)$ $P(E)=\frac{4}{5}, P(F)=\frac{3}{5}$
$(B)$ $P(E)=\frac{1}{5}, P(F)=\frac{2}{5}$
$(C)$ $P(E)=\frac{2}{5}, P(F)=\frac{1}{5}$
$(D)$ $P(E)=\frac{3}{5}, P(F)=\frac{4}{5}$

$A$ 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 $P(E \cup F)$.