An ellipse is inscribed in a circle and a point is chosen at random inside the circle. If the probability that this point lies outside the ellipse is $\frac{2}{3}$,then the eccentricity of the ellipse is $\frac{a\sqrt{b}}{c}$,where $\gcd(a, c) = 1$ and $b$ is a square-free integer. Find the value of $a \cdot b \cdot c$.

Let $S = \{w_1, w_2, \ldots\}$ be the sample space associated with a random experiment. Let $P(w_n) = \frac{P(w_{n-1})}{2}$ for $n \geq 2$. Let $A = \{2k + 3\ell : k, \ell \in \mathbb{N}\}$ and $B = \{w_n : n \in A\}$. Then $P(B)$ is equal to:

$A$ 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 $P(A') + P(B') P(A \cup B) = 0.7$,then $P(A') + P(B')$ is

Football teams $T_1$ and $T_2$ play two games against each other. The outcomes of the two games are independent. The probabilities of $T_1$ winning,drawing,and losing a game against $T_2$ are $\frac{1}{2}$,$\frac{1}{6}$,and $\frac{1}{3}$,respectively. Each team gets $3$ points for a win,$1$ point for a draw,and $0$ points for a loss. Let $X$ and $Y$ denote the total points scored by teams $T_1$ and $T_2$,respectively,after two games.
$(1)$ $P(X>Y)$ is
$(A)$ $\frac{1}{4}$ $(B)$ $\frac{5}{12}$ $(C)$ $\frac{1}{2}$ $(D)$ $\frac{7}{12}$
$(2)$ $P(X=Y)$ is
$(A)$ $\frac{11}{36}$ $(B)$ $\frac{1}{3}$ $(C)$ $\frac{13}{36}$ $(D)$ $\frac{1}{2}$

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\}) = $