Find $n(A \cap B)$ Question:
In group of people, 60% like coffee and 70% like tea. How many people like both of them.?
My Effort:
We have to find how many people like both the items that means we have to find $n(A \cap B)$.
So, I have assume that $n(A)= 60$, $n(B)=70$ and $n(U)=100$.
Normally these type problem can be done using formula $n(A \cup B)= n(A)+n(B)-n(A \cap B)$. But here $n(A \cup B)$ is not given.
p.s. Answer given is $30 \le x \le 60.$
 A: You need to use some 'common sense' ideas to give you more than just the equation $n(A\cup B)=n(A)+n(B)-n(A\cap B)$. 
The first is that $n(A\cup B)\leq 100$. This gives you that $n(A\cap B)\geq n(A)+n(B)-100=60+70-100=30$.
The second is that $n(A\cap B)\leq n(B)=60$.
This gives you the range $30\leq n(A\cap B)\leq 60$.
You then need to check that everything in this range is in fact possible. Ask yourself 'given $x$ in the range 30 to 60, is it possible that $n(A\cap B)=x$?' The answer to this is 'yes'.
A: Let total number of people be $n$, thenpeople liking coffee is $0.6n$people liking tea is $0.7n$let people liking both be $xn$
$n(A \cup B)= n(A)+n(B)-n(A \cap B)$, gives
$n=(0.6+0.7)n-xn \Rightarrow x=0.3$, this is when $A, B$ do not overlap completely, if they overlap completely then $x=0.6$
A: For the maximum of $P(C \cap T)$ I would calculate $Max(P(C),P(T))$
And for minimum of $P(C \cap T)$ I would calculate $P(\overline C\cap T)=Min(P(\overline C),P(T)) =A$ and  $P(\overline T\cap C)=Min(P(\overline T),P(C))=B$. Then the Minimum of $P(C \cap T)$ is $1-(A+B)$.
A: You are almost there:
$$
70=\max\{n(A),n(B)\}\leq n(A\cup B)\leq n(U)=100
$$
so
$$
n(A\cap B)=n(A)+n(B)-n(A\cup B)=130-n(A\cup B)\in[130-100,130-70]=[30,60].
$$
