$A$ cubical die with faces marked $1, 2, 3, ..., 6$ is tossed such that the probability of throwing the number $t$ is proportional to $t^2$. The probability that the number $5$ has appeared,given that the number turned up is not even,is:

$A$ biased die is tossed and the respective probabilities for various faces to turn up are given below:

$Face$	$1$	$2$	$3$	$4$	$5$	$6$
$P(F)$	$0.1$	$0.24$	$0.19$	$0.18$	$0.15$	$0.14$

If an even face has turned up,then the probability that it is face $2$ or face $4$ is:

Let $A$ and $B$ be independent events with $P(A)=0.3$ and $P(B)=0.4$. Find $P(B | A)$.

Two integers $r$ and $s$ are drawn one at a time without replacement from the set $\{1, 2, \ldots, n\}$. Then $P(r \leq k \mid s \leq k) =$

If $A$ and $B$ are two events such that $P(A) = 0.7$,$P(B) = 0.4$,and $P(A \cap \overline{B}) = 0.5$,where $\overline{B}$ denotes the complement of $B$,then $P(B \mid (A \cup \overline{B}))$ is equal to:

Bag $A$ contains $9$ white and $8$ black balls, while bag $B$ contains $6$ white and $4$ black balls. One ball is randomly picked up from bag $B$ and mixed with the balls in bag $A$. Then a ball is randomly drawn from bag $A$. If the probability that the ball drawn is white is $p/q$ (where $gcd(p,q)=1$), then $p+q$ is equal to: