$A$ and $B$ are two events such that $P(A) = 0.8$,$P(B) = 0.6$,and $P(A \cap B) = 0.5$. Then the value of $P(A/B)$ is:

Let $A$ and $E$ be any two events with positive probabilities:
Statement $- 1$: $P(E/A) \geq P(A/E)P(E)$
Statement $- 2$: $P(A/E) \geq P(A \cap E)$

$3$ cards are drawn one-by-one without replacement from a well-shuffled pack of $52$ cards. The probability that the first card is a heart,the second is a queen,and the third is a king is equal to:

If $A$ and $B$ are two events of a sample space $S$ such that $P(A)=0.2$,$P(B)=0.6$ and $P(A \mid B)=0.5$,then $P(A^{\prime} \mid B) = $

Given two independent events $A$ and $B$ such that $P(A) = 0.3$ and $P(B) = 0.6$. Find $P(A \text{ and not } B)$.

If $A$ and $B$ are two events such that $P(A | B) = 0.6$, $P(B | A) = 0.3$, and $P(A) = 0.1$, then $P(\bar{A} \cap \bar{B})$ equals: