$A$ bag contains $19$ red balls and $19$ black balls. Two balls are chosen at a time repeatedly and discarded if they are of the same colour,but if they are different,the black ball is discarded and the red ball is returned to the bag. The probability that this process will terminate with one red ball is

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:

If a $3$-digit number is randomly chosen,what is the probability that either the number itself or some permutation of the number (which is a $3$-digit number) is divisible by $4$ and $5$?

If $12$ identical balls are to be placed randomly in $3$ identical boxes,then the probability that one of the boxes contains exactly $3$ balls is

Two fair dice,each with faces numbered $1, 2, 3, 4, 5$ and $6$,are rolled together and the sum of the numbers on the faces is observed. This process is repeated until the sum is either a prime number or a perfect square. Suppose the sum turns out to be a perfect square before it turns out to be a prime number. If $p$ is the probability that this perfect square is an odd number,then the value of $14p$ is . . . . .