One die has two faces marked $1$,two faces marked $2$,one face marked $3$ and one face marked $4$. Another die has one face marked $1$,two faces marked $2$,two faces marked $3$ and one face marked $4$. The probability of getting the sum of numbers to be $4$ or $5$,when both the dice are thrown together,is

Two players $A$ and $B$ alternatively toss $3$ coins simultaneously. The player who gets $2$ heads and $1$ tail first, wins the game. If the game continues until someone wins and if $A$ begins the game, the probability that $B$ wins the game is

Three dice are thrown simultaneously and the sum of the numbers appeared on them is noted. If $A$ is the event of getting a sum greater than $14$ and $B$ is the event of getting a sum which is a multiple of $3$,then $P(A \cap \overline{B}) + P(\overline{A} \cap B) = $

If four positive integers are selected randomly from the set of positive integers,then the probability that the unit digit of their product is $1, 3, 7,$ or $9$ 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\}) = $

$A$ and $B$ are independent events. The probability that both $A$ and $B$ occur is $\frac{1}{20}$ and the probability that neither of them occurs is $\frac{3}{5}$. The probability of occurrence of $A$ is