Let $E$ and $F$ be two independent events. The probability that both $E$ and $F$ happen is $\frac{1}{12}$ and the probability that neither $E$ nor $F$ happens is $\frac{1}{2}$. Then a value of $\frac{P(E)}{P(F)}$ is

$A$ 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 $r$)

$A$ player $X$ has a biased coin whose probability of showing heads is $p$ and a player $Y$ has a fair coin. They start playing a game with their own coins and play alternately. The player who throws a head first is a winner. If $X$ starts the game,and the probability of winning the game by both the players is equal,then the value of $p$ is

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:

The probability that a mechanic makes an error while using a machine on the $n$th day is given by $P(E_n) = \frac{1}{2^n}$. If he has operated the machine for $4$ days, the probability that he has not made a mistake on $3$ of the $4$ days 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\}) = $