Consider the system of equations $$ax+by=0; cx+dy=0,$$
where $$a,b,c,d$$ $$ \in \left\{ {0,1} \right\}$$

STATEMENT - 1 : The probability that the system of equations has a unique solution is $${3 \over 8}.$$ and

STATEMENT - 2 : The probability that the system of equations has a solution is $$1.$$

Let $${H_1},{H_2},....,{H_n}$$ be mutually exclusive and exhaustive events with $$P\left( {{H_1}} \right) > 0,i = 1,2,.....,n.$$ Let $$E$$ be any other event with $$0 < P\left( E \right) < 1.$$
STATEMENT-1:
$$P\left( {{H_1}|E} \right) > P\left( {E|{H_1}} \right).P\left( {{H_1}} \right)$$ for $$i=1,2,....,n$$ because

STATEMENT-2: $$\sum\limits_{i = 1}^n {P\left( {{H_i}} \right)} = 1.$$

One Indian and four American men and their wives are to be seated randomly around a circular table. Then the conditional probability that the Indian man is seated adjacent to his wife given that each American man is seated adjacent to his wife is

Let $${E^c}$$ denote the complement of an event $$E.$$ Let $$E, F, G$$ be pairwise independent events with $$P\left( G \right) > 0$$ and $$P\left( {E \cap F \cap G} \right) = 0.$$ Then $$P\left( {{E^c} \cap {F^c}|G} \right)$$ equals