Let $A = \{a_1, a_2, a_3, \dots, a_n\}$ be a set containing $n$ elements. Two subsets $P$ and $Q$ of $A$ are formed independently. The number of ways in which these subsets can be formed such that $(P - Q)$ contains exactly $2$ elements is:

An electronic assembly consists of two subsystems $A$ and $B$. Past testing procedures data show that probabilities of failure are $P(A \text{ fails}) = 0.2$, $P(B \text{ fails alone}) = 0.15$, and $P(A \cap B \text{ fail}) = 0.15$. Then the probability that $A$ fails alone is equal to

If $P(A^c) = 0.3$,$P(B) = 0.4$,and $P(A \cap B^c) = 0.5$,then find $P[B / (A \cup B)^c]$.

In a certain examination,a candidate has to pass in each of the $5$ subjects. The number of ways in which the candidate can fail is:

In a town of $10,000$ families,it was found that $40\%$ of families buy newspaper $A$,$20\%$ buy newspaper $B$,and $10\%$ buy newspaper $C$. Also,$5\%$ of families buy $A$ and $B$,$3\%$ buy $B$ and $C$,and $4\%$ buy $A$ and $C$. If $2\%$ of families buy all three newspapers,then the number of families that buy newspaper $A$ only is:

Two newspapers $A$ and $B$ are published in a city. It is known that $25\%$ of the city population reads $A$ and $20\%$ reads $B$,while $8\%$ reads both $A$ and $B$. Further,$30\%$ of those who read $A$ but not $B$ look into advertisements,$40\%$ of those who read $B$ but not $A$ look into advertisements,and $50\%$ of those who read both $A$ and $B$ look into advertisements. The percentage of the population who look into advertisements is: