The probability that $A$ speaks truth is $75 \%$ and the probability that $B$ speaks truth is $80 \%$. The probability that they contradict each other when asked to speak on a fact is

Let there be three independent events $E_{1}, E_{2}$ and $E_{3}$. The probability that only $E_{1}$ occurs is $\alpha$,only $E_{2}$ occurs is $\beta$ and only $E_{3}$ occurs is $\gamma$. Let $p$ denote the probability that none of the events occur,which satisfies the equations $(\alpha - 2\beta)p = \alpha\beta$ and $(\beta - 3\gamma)p = 2\beta\gamma$. All the given probabilities are assumed to lie in the interval $(0, 1)$. Then,$\frac{\text{Probability of occurrence of } E_{1}}{\text{Probability of occurrence of } E_{3}}$ is equal to ..........

Two players $A$ and $B$ are alternately throwing a coin and a die together. $A$ player who first throws a head and a $6$ wins the game. If $A$ starts the game, then the probability that $B$ wins the game is:

Let $|X|$ denote the number of elements in set $X$. Let $S = \{1, 2, 3, 4, 5, 6\}$ be a sample space,where each element is equally likely to occur. If $A$ and $B$ are independent events associated with $S$,then the number of ordered pairs $(A, B)$ such that $1 \leq |B| < |A|$ equals:

If $E$ and $F$ are independent events such that $0 < P(E) < 1$ and $0 < P(F) < 1,$ then

Let the mean and variance of $7$ observations $2, 4, 10, x, 12, 14, y$ (where $x > y$) be $8$ and $16$ respectively. Two numbers are chosen from the set $\{1, 2, 3, x-4, y, 5\}$ one after another without replacement. Then the probability that the smaller number among the two chosen numbers is less than $4$ is: