$U_1, U_2, U_3$ are three urns. $U_1$ contains $5$ red,$3$ white,$2$ black balls; $U_2$ contains $4$ red,$4$ white,$2$ black balls and $U_3$ contains $3$ red,$4$ white,$3$ black balls. If a ball is chosen at random from an urn chosen at random,then the probability of not getting a black ball is

$A$ bag contains $19$ red balls and $19$ black balls. Two balls are chosen at a time repeatedly and discarded if they are of the same colour,but if they are different,the black ball is discarded and the red ball is returned to the bag. The probability that this process will terminate with one red ball is

Suppose four balls labelled $1, 2, 3, 4$ are randomly placed in boxes $B_1, B_2, B_3, B_4$. The probability that exactly one box is empty is

Cards are drawn one after the other without replacement from a well-shuffled pack of cards until an ace card appears. If the probability that exactly $5$ cards are drawn before the first ace card appears is $\frac{4}{49}\left(\frac{p_1 \cdot p_2 \cdot p_3}{p_4 \cdot p_5 \cdot p_6}\right)$, where $p_i$ is prime for $i=1, 2, 3, 4, 5, 6$, then $(\max \{p_i\} - \min \{p_i\}) = $

If $A$ and $B$ are two independent events such that $P(A) > 0.5$,$P(B) > 0.5$,$P(A \cap \bar{B}) = \frac{3}{25}$,and $P(\bar{A} \cap B) = \frac{8}{25}$,then $P(A \cap B)$ is:

If $A$ and $B$ are two independent events such that $P(B)=\frac{2}{7}$ and $P\left(A \cup B^c\right)=0.8$, then $P(A \cup B)$ $=$