Let $A$ be the set of all $4$-digit natural numbers whose exactly one digit is $7$. Then the probability that a randomly chosen element of $A$ leaves a remainder of $2$ when divided by $5$ 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:

$A$ bag contains $4$ red and $6$ black balls. $A$ ball is drawn at random from the bag,its colour is observed,and this ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag,then the probability that this drawn ball is red,is:

Two natural numbers are chosen at random from $1$ to $100$ and are multiplied. If $A$ is the event that the product is an even number and $B$ is the event that the product is divisible by $4$,then $P(A \cap \bar{B})=$

Let $A, B, C$ be three pairwise independent events of a random experiment. If $P(\bar{B} \cup \bar{C}) = \frac{1}{2}$, $P(A) > 0$, $P(B) = b$, and $P(C) = c$, then $P((\bar{B} \cap \bar{C}) \mid A) = $

One die has two faces marked $1$,two faces marked $2$,one face marked $3$ and one face marked $4$. Another die has one face marked $1$,two faces marked $2$,two faces marked $3$ and one face marked $4$. The probability of getting the sum of numbers to be $4$ or $5$,when both the dice are thrown together,is