$A$ and $B$ alternately throw a pair of dice. A wins if he throws a sum of 5 before $B$ throws a sum of 8 , and $B$ wins if he throws a sum of 8 before $A$ throws a sum of 5 . The probability, that A wins if A makes the first throw, is

A board has 16 squares as shown in the figure :

Out of these 16 squares, two squares are chosen at random. The probability that they have no side in common is :

One die has two faces marked 1 , two faces marked 2 , one face marked 3 and one face marked 4 . Another die has one face marked 1 , two faces marked 2 , two faces marked 3 and one face marked 4. The probability of getting the sum of numbers to be 4 or 5 , when both the dice are thrown together, is

If $A$ and $B$ are two events such that $P(A \cap B)=0.1$, and $P(A \mid B)$ and $P(B \mid A)$ are the roots of the equation $12 x^2-7 x+1=0$, then the value of $\frac{P(\bar{A} \cup \bar{B})}{P(\bar{A} \cap \bar{B})}$ is :