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

Let $A$ and $B$ be two events with $P(A^{C}) = 0.3$,$P(B) = 0.4$,and $P(A \cap B^{C}) = 0.5$. Then $P(B \mid A \cup B^{C})$ is equal to

If $A$ and $B$ are any two events such that $P(A) = \frac{2}{5}$ and $P(A \cap B) = \frac{3}{20}$,then the conditional probability $P(A | A' \cup B')$,where $A'$ denotes the complement of $A$,is equal to:

If $P(AB) = P(A)P(B)$,$P(A/B) = 1/4$,and $P(B/A) = 1/3$,then which of the following is true?

If $A$ and $B$ are events,such that $P(A) = \frac{1}{4}$,$P(A|B) = \frac{1}{2}$,and $P(B|A) = \frac{2}{3}$,then $P(B)$ is

Let $E_{1}$ and $E_{2}$ be two events such that the conditional probabilities $P(E_{1} \mid E_{2}) = \frac{1}{2}$,$P(E_{2} \mid E_{1}) = \frac{3}{4}$ and $P(E_{1} \cap E_{2}) = \frac{1}{8}$. Then: