Let $$\mathrm{A}$$ and $$\mathrm{B}$$ be two events such that $$P(B \mid A)=\frac{2}{5}, P(A \mid B)=\frac{1}{7}$$ and $$P(A \cap B)=\frac{1}{9} \cdot$$ Consider

(S1) $$P\left(A^{\prime} \cup B\right)=\frac{5}{6}$$,

(S2) $$P\left(A^{\prime} \cap B^{\prime}\right)=\frac{1}{18}$$

Then :

Out of $$60 \%$$ female and $$40 \%$$ male candidates appearing in an exam, $$60 \%$$ candidates qualify it. The number of females qualifying the exam is twice the number of males qualifying it. A candidate is randomly chosen from the qualified candidates. The probability, that the chosen candidate is a female, is :

Let X have a binomial distribution B(n, p) such that the sum and the product of the mean and variance of X are 24 and 128 respectively. If $$P(X>n-3)=\frac{k}{2^{n}}$$, then k is equal to :

A six faced die is biased such that

$$3 \times \mathrm{P}($$a prime number$$)\,=6 \times \mathrm{P}($$a composite number$$)\,=2 \times \mathrm{P}(1)$$.

Let X be a random variable that counts the number of times one gets a perfect square on some throws of this die. If the die is thrown twice, then the mean of X is :