$A$ bag $P$ contains $4$ red and $5$ black balls,another bag $Q$ contains $3$ red and $6$ black balls. If one ball is drawn at random from bag $P$ and two balls are drawn from bag $Q$,then the probability that out of the three balls drawn two are black and one is red,is

Let $0 < P(A) < 1$,$0 < P(B) < 1$ and $P(A \cup B) = P(A) + P(B) - P(A)P(B).$ Then

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:

Two dice are thrown and two coins are tossed simultaneously. The probability of getting prime numbers on both the dice along with a head and a tail on the two coins is

For independent events $A$ and $B$,if $P(A) = \frac{1}{2}$ and $P(A \cup B) = \frac{3}{5}$,then $P(B) =$ . . . . . . .

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?
$(A)$ $P(E)=\frac{4}{5}, P(F)=\frac{3}{5}$
$(B)$ $P(E)=\frac{1}{5}, P(F)=\frac{2}{5}$
$(C)$ $P(E)=\frac{2}{5}, P(F)=\frac{1}{5}$
$(D)$ $P(E)=\frac{3}{5}, P(F)=\frac{4}{5}$