$A$ fair die is rolled indefinitely. Player $A$ wins if two consecutive rolls show $3$ or $5$, and player $B$ wins if two consecutive rolls show $1$ or $2$ or $4$ or $6$. The probability that player $A$ wins is:

Two fair dice,each with faces numbered $1, 2, 3, 4, 5$ and $6$,are rolled together and the sum of the numbers on the faces is observed. This process is repeated until the sum is either a prime number or a perfect square. Suppose the sum turns out to be a perfect square before it turns out to be a prime number. If $p$ is the probability that this perfect square is an odd number,then the value of $14p$ 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\}) = $

$A$ random variable $X$ has the following probability distribution:
| $X=x$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| $P(X=x)$ | $0.15$ | $0.23$ | $0.12$ | $0.20$ | $0.08$ | $0.10$ | $0.05$ | $0.07$ |
For the events $E = \{X \text{ is a prime number}\}$ and $F = \{X < 5\}$,find $P(E \cup F)$.

If $E_1, E_2, \ldots, E_n$ are independent events such that $P(E_r) = \frac{1}{1+r}$ for $r = 1, 2, \ldots, n$, then the probability that at least one of $E_1, E_2, \ldots, E_n$ happens is

$A$ bag contains $2$ white,$3$ green,and $5$ red balls. If three balls are drawn one after the other without replacement,then the probability that the last ball drawn was red is