For independent events $A$ and $B$,$P(A \cup B) =$ . . . . . . .

Two squares are chosen one by one on a chessboard. The probability that they have a side in common is

The probabilities that players $A$ and $B$ of a team are selected for the captaincy for a tournament are $0.6$ and $0.4$,respectively. If $A$ is selected as the captain,the probability that the team wins the tournament is $0.8$ and if $B$ is selected as the captain,the probability that the team wins the tournament is $0.7$. Then the probability,that the team wins the tournament,is:

If $A$ and $B$ are two events,then the probability of the event that at most one of $A$ and $B$ occurs 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:

There are three bags $B_1$,$B_2$,and $B_3$ containing $2$ Red and $3$ White,$5$ Red and $5$ White,and $3$ Red and $2$ White balls respectively. $A$ ball is drawn from bag $B_1$ and placed in bag $B_2$,then a ball is drawn from bag $B_2$ and placed in bag $B_3$,then a ball is drawn from bag $B_3$. The number of ways in which this process can be completed,if the same colour balls are used in the first and second transfers (assume all balls to be distinct),is