An urn $A$ contains $3$ white and $5$ black balls. Another urn $B$ contains $6$ white and $8$ black balls. $A$ ball is picked from $A$ at random and then transferred to $B$. Then, a ball is picked at random from $B$. The probability that it is a white ball is

$A$ candidate takes three tests in succession and the probability of passing the first test is $p$. The probability of passing each succeeding test is $p$ if he passes the preceding one, and $\frac{p}{2}$ if he fails the preceding one. The candidate is selected if he passes at least two tests. The probability that the candidate is selected is:

Two persons $P$ and $Q$ are considering applying for a job. The probability that $P$ applies for the job is $1/4$,the probability that $P$ applies for the job given that $Q$ applies for the job is $1/2$,and the probability that $Q$ applies for the job given that $P$ applies for the job is $1/3$. Then the probability that $P$ does not apply for the job given that $Q$ does not apply for the job is

If $A$ and $B$ are two independent events such that $P(A \cap B') = \frac{3}{25}$ and $P(A' \cap B) = \frac{8}{25}$,then $P(A) = $

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

If $P(E)=0.8, P(F)=0.5$ and $P(F \mid E)=0.4$ then,$P(E \mid F)=$ . . . . . . .