Four persons $A, B, C$ and $D$ throw an unbiased die, turn by turn in succession till one gets an even number and wins the game. What is the probability that $A$ wins the game if $A$ begins?

If $E$ and $F$ are events with $P(E) \le P(F)$ and $P(E \cap F) > 0,$ then

Of the 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 the probability $p$ that none of events $E_1, E_2$ or $E_3$ occurs satisfy 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} = $

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

Consider a system of equations $ax + by = 0$ and $cx + dy = 0$,where $a, b, c, d \in \{0, 1\}$.
Statement $-1$: The probability that the system of equations has a solution is $1$.
Statement $-2$: The probability that the system of equations has a unique solution is $\frac{3}{8}$.

$A$ locker can be opened by dialing a fixed three-digit code (between $000$ and $999$). $A$ stranger who does not know the code tries to open the locker by dialing three digits at random. The probability that the stranger succeeds at the $k^{th}$ trial is