$A$ bag contains $2n$ coins,out of which $n-1$ are unfair with heads on both sides and the remaining are fair. One coin is picked from the bag at random and tossed. If the probability that a head appears in the toss is $\frac{41}{56}$,then the number of unfair coins in the bag is:

Let $B_{i} (i=1, 2, 3)$ be three independent events in a sample space. The probability that only $B_{1}$ occurs is $\alpha$,only $B_{2}$ occurs is $\beta$,and only $B_{3}$ occurs is $\gamma$. Let $p$ be the probability that none of the events $B_{i}$ occurs,and these $4$ probabilities satisfy the equations $(\alpha - 2\beta)p = \alpha\beta$ and $(\beta - 3\gamma)p = 2\beta\gamma$ (All the probabilities are assumed to lie in the interval $(0, 1)$). Then $\frac{P(B_{1})}{P(B_{3})}$ is equal to ..........

An urn $A$ contains $4$ white and $1$ black ball; urn $B$ contains $3$ white and $2$ black balls and urn $C$ contains $2$ white and $3$ black balls. One ball is transferred randomly from $A$ to $B$; later one ball is transferred randomly from $B$ to $C$. Finally,if a ball is drawn randomly from $C$,then the probability that it is a black ball is

$A$ and $B$ throw a pair of dice alternately and they note the sum of the numbers appearing on the dice. $A$ wins if he throws $6$ before $B$ throws $7$,and $B$ wins if he throws $7$ before $A$ throws $6$. If $A$ begins,the probability of $A$ winning is:

$A$ bag contains $4$ red and $6$ black balls. $A$ ball is drawn at random from the bag,its colour is observed,and this ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag,then the probability that this drawn ball is red,is:

$A$ coin is tossed $m + n$ times,where $m \ge n.$ The probability of getting at least $m$ consecutive heads is