Given that $E$ and $F$ are events such that $P(E)=0.6$,$P(F)=0.3$,and $P(E \cap F)=0.2$,find $P(E|F)$ and $P(F|E)$.

$A$ 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:

Given $P(A)=0.5, P(B)=0.4, P(A \cap B)=0.3$,then $P(A^{\prime} / B^{\prime})$ is equal to

If $P(A | B) > P(A)$,then which of the following is correct?

Let $X$ be a random variable which takes values $1, 2, 3, 4$ such that $P(X=r) = K r^3$ where $r = 1, 2, 3, 4$. Then:

Two cards are drawn one by one from a pack of cards. The probability of getting the first card as an ace and the second as a coloured card is (without replacement).