The probabilities of three events $A, B$ and $C$ are given by $P(A)=0.6, P(B)=0.4$ and $P(C)=0.5$. If $P(A \cup B)=0.8, P(A \cap C)=0.3, P(A \cap B \cap C)=0.2, P(B \cap C)=\beta$ and $P(A \cup B \cup C)=\alpha$,where $0.85 \leq \alpha \leq 0.95$,then $\beta$ lies in the interval:

There are $100$ students in a class. In an examination,$50$ of them failed in Mathematics,$45$ failed in Physics,$40$ failed in Biology and $32$ failed in exactly two of the three subjects. Only one student passed in all the subjects. Then,the number of students failing in all the three subjects is:

In a class of $100$ students,$15$ students chose only physics (but not mathematics and chemistry),$3$ chose only chemistry (but not mathematics and physics),and $45$ chose only mathematics (but not physics and chemistry). Of the remaining students,it is found that $23$ have taken physics and chemistry,$20$ have taken physics and mathematics,and $12$ have taken mathematics and chemistry. The number of students who chose all the three subjects is

Out of $60 \%$ female and $40 \%$ male candidates appearing in an exam,$60 \%$ of the total 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 $A = \{n \in N : n \text{ is a } 3 \text{-digit number}\}$. Let $B = \{9k + 2 : k \in N\}$ and $C = \{9k + l : k \in N\}$ for some $l$ $(0 < l < 9)$. If the sum of all the elements of the set $A \cap (B \cup C)$ is $274 \times 400$,then $l$ is equal to:

In a certain town,$25\%$ of the families own a phone and $15\%$ own a car; $65\%$ families own neither a phone nor a car and $2,000$ families own both a car and a phone. Consider the following three statements:
$(A) \, 5\%$ families own both a car and a phone
$(B) \, 35\%$ families own either a car or a phone
$(C) \, 40,000$ families live in the town
Then,