Inclusion-exclusion principle of professors liking $3$ different games Suppose $60$% of all college professors like tennis, $65$% like bridge, and $50$% like chess; $45$% like any given pair of recreations:a) Should you be suspicious if told that $20$% like all three recreations?b) What is the smallest percentage who could like all three recreactions?c) What is the largest percentage who could like all three recreactions?
My thinking so far is assuming there are $100$ professors for simplicity, and show what set information I have, which is:$$N = 100$$$$|T|=60$$$$|B|=65$$$$|C|=50$$$$|T\cap B+B\cap C+C\cap T|=45$$$$|T\cap \overline {(T\cap B)\cup (T\cap C)}+B\cap \overline {(B\cap C)\cup (B\cap T)}+C\cap \overline {(C\cap T)\cup (C\cap B)}-T\cup B \cup C|=55$$
But from here I don't know how to arrive at $|T\cap B\cap C|$. Any suggestions?
 A: You misinterpreted the statement that $45\%$ like any given pair of recreations.  With your assumption that there are $100$ professors, it means that $|T \cap B| = |T \cap C| = |B \cap C| = 45$.  
By the Inclusion-Exclusion Principle, 
$$|T \cup B \cup C| = |T| + |B| + |C| - |T \cap B| - |T \cap C| - |B \cap C| + |T \cap B \cap C| \tag{1}$$
Notice that your assumption that there are $100$ professors means that 
$$|T \cup B \cup C| \leq 100 \tag{2}$$
Since $65\%$ of these hundred professors enjoy bridge, $|B| = 65$.  Thus, 
$$|T \cup B \cup C| \geq 65 \tag{3}$$
Substituting $60$ for $|T|$, $65$ for $|B|$, $50$ for $|C|$, $45$ for $|T \cap B|$, $45$ for $|T \cap C|$, and $45$ for $|B \cap C|$ in equation 1 yields
$$|T \cup B \cup C| = 60 + 65 + 50 - 45 - 45 - 45 + |T \cap B \cap C| = 40 + |T \cap B \cap C| \tag{4}$$
Comparing equation 4 with inequalities 2 and 3 imposes the constraints

 $$25 \leq |T \cap B \cap C| \leq 60$$

Notice also that 
\begin{align*}
T \cap B \cap C & \subseteq T \cap B\\
T \cap B \cap C & \subseteq T \cap C\\
T \cap B \cap C & \subseteq B \cap C
\end{align*}
Since $|T \cap B| = |T \cap C| = |B \cap C| = 45$, we have the additional constraint 

 $$|T \cap B \cap C| \leq 45$$

By considering those constraints, you can answer the three questions you posed above. 
A: I have read the line 45% like any given pair, as a statement that they might like the 3rd.
Lets start by bucketing the profs into like 1, like 2, like 3.
for a moment lets assume that profs only like 1.
60+65+50 = 175... that makes 75 too many.  But as we put a prof into the like 2 bucket we can remove them from 2 of the like 1 buckets.  And if we allocate a prof to the like 3 bucket we can remove them from all 3 like 1 buckets.
Like 1 + Like 2 + Like 3 = 100
Like 2 + Like 3 = 45
like 1 = 55
like 2 + 2*Like 3 = 75
like 3 = 30
Now it is possible that I have misread it, and if you like any given pair, you like only that pair, and not the 3rd.
in which case:
Like 1 + Like 2 + Like 3 = 100
Like 2 = 45
like 2 + 2*Like 3 = 75
like 3 = 15
