If $P(A)=0.25, P(B)=0.55$ and $P(A \cup B)=0.65$,then $P(B' \mid A) =$ . . . . . . .

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 $4\,P(A) = 6\,P(B) = 10\,P(A \cap B) = 1$,then $P\left( \frac{B}{A} \right) = \dots$

Let $A$ and $B$ be two events such that $P(A) = \frac{3}{8}$, $P(B) = \frac{5}{8}$ and $P(A \cup B) = \frac{3}{4}$. Then $P(A'|B) - P(A|B) =$ . . . . . .

An electronic assembly consists of two subsystems,$A$ and $B$. From previous testing procedures,the following probabilities are known:
$P(A \text{ fails}) = 0.2$
$P(B \text{ fails alone}) = 0.15$
$P(A \text{ and } B \text{ fail}) = 0.15$
Evaluate the probability $P(A \text{ fails } | \text{ } B \text{ has failed})$.

If $P(A)=0.8, P(B)=0.5$ and $P(B | A)=0.4,$ find $P(A \cap B).$