Three numbers are selected at random from the set ${1, 2, 3, \dots, 8}$ without replacement. Given that the minimum of the selected numbers is $3$ and the maximum is $6$, what is the probability that the third number is $4$ or $5$?

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} = $

$A$ box contains coupons labelled $1, 2, 3, \ldots, n$. $A$ coupon is picked at random and the number $x$ is noted. The coupon is put back into the box and a new coupon is picked at random. The new number is $y$. Then,the probability that one of the numbers $x, y$ divides the other is (in the options below $[r]$ denotes the largest integer less than or equal to $r$)

There are four machines and it is known that exactly two of them are faulty. They are tested one by one, in a random order until both the faulty machines are identified. The probability that only two tests are needed is:

Three numbers are chosen at random,one after another with replacement,from the set $S = \{1, 2, 3, \ldots, 100\}$. Let $p_1$ be the probability that the maximum of the chosen numbers is at least $81$ and $p_2$ be the probability that the minimum of the chosen numbers is at most $40$.
$(1)$ The value of $\frac{625}{4} p_1$ is
$(2)$ The value of $\frac{125}{4} p_2$ is

If two cards are drawn at random simultaneously from a pack of $52$ playing cards, then the probability of getting a face card and a spade card other than the face card is: