 Solution

For finite sets $A_1,\dots ,A_s$, the inclusion-exclusion principle statesTo prove it, count the contribution of an element lying in exactly $r\ge 1$ of the sets. Its total coefficient isby the binomial theorem. An element in none contributes zero. Summing these elementwise counts proves the formula. Applying the same argument to indicator functions and taking expectations proves the corresponding probability formula.

Factor $4199=13\cdot 17\cdot 19$. An integer is coprime to this product precisely when none of its three prime factors divides it. Inclusion-exclusion applied to the multiples of those prime numbers among $1,\dots ,4199$ gives Euler's totient functionFor the survey, let $H,D,S$ be the three detestation events and use proportions as probabilities. The union $H\cup D$ is the reported total union with the only-$S$ group removed, so $\mathbb{P}(H\cup D)=0.90-0.27=0.63$. The two-set inclusion-exclusion principle gives $\mathbb{P}(H\cap D)=0.45+0.28-0.63=0.10$. Removing the all-three group leavesThe required proportion is 4%. The separately reported total $S$ proportion is consistent with, but unnecessary for, this calculation.