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 $E_1$ and $E_2$ are two events of a sample space such that $P(E_1) = \frac{1}{4}$,$P(E_2 \mid E_1) = \frac{1}{2}$,and $P(E_1 \mid E_2) = \frac{1}{4}$,then $P(\bar{E}_1 \mid E_2) = $

If $A$ and $B$ are two events such that $P(A) = \frac{1}{3}$,$P(B) = \frac{1}{5}$,and $P(A \cup B) = \frac{1}{3}$,then the value of $P(A^{\prime} | B^{\prime}) + P(B^{\prime} | A^{\prime})$ is:

$A$ die is thrown twice and the sum of the numbers appearing is observed to be $6$. What is the conditional probability that the number $4$ has appeared at least once?

If $E_1$ and $E_2$ are two events of the sample space such that $P(E_1) = \frac{1}{4}$,$P(E_1 | E_2) = \frac{1}{2}$ and $P(E_2 | E_1) = \frac{1}{3}$,then $P(E_1 | \bar{E}_2) = $

Compute $P(A | B),$ if $P(B)=0.5$ and $P(A \cap B)=0.32$.