For events $A_1,\ldots,A_n$, the probabilistic <inclusion-exclusion principle> is
$$
\boxed{\mathbb P\left(\bigcup_{i=1}^nA_i\right)
=\sum_{\varnothing\ne J\subseteq\{1,\ldots,n\}}
(-1)^{|J|+1}\mathbb P\left(\bigcap_{j\in J}A_j\right)}.
$$
To prove it, fix an outcome lying in exactly $r$ events. Its total coefficient on the right is
$$
\sum_{k=1}^r(-1)^{k+1}\binom rk=1
$$
by the <binomial theorem>; outcomes in no event contribute zero. Taking expectations of this pointwise indicator identity proves the formula. Truncating after the pair terms gives the <Bonferroni inequalities>
$$
\mathbb P\left(\bigcup_iA_i\right)
\geq\sum_i\mathbb P(A_i)-\sum_{i<j}\mathbb P(A_i\cap A_j).
$$

For the final bound, let $X=\sum_i\mathbf1_{A_i}$ and $S=\mathbb EX=\sum_i\mathbb P(A_i)$. The intersection assumption gives
$$
\mathbb E[X^2]
=S+2\sum_{i<j}\mathbb P(A_i\cap A_j)
\leq S+n-1.
$$
The <Cauchy-Schwarz inequality> applied to $X\mathbf1_{X>0}$ yields $S^2\leq\mathbb E[X^2]\mathbb P(X>0)\leq S+n-1$. Therefore
$$
S\leq\frac{1+\sqrt{4n-3}}2\leq2\sqrt n,
$$
so the required universal constant may be taken as $\boxed{c=2}$. This is a <second moment method> estimate.

Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ia / Paper 1 / 11F / Solution

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{wiki}

= Bonferroni inequality
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Truncating the <inclusion-exclusion principle> after an odd number of terms gives an upper bound, and truncating after an even number gives a lower bound. In particular,
$$
\mathbb P\left(\bigcup_iA_i\right)
\geq\sum_i\mathbb P(A_i)-\sum_{i<j}\mathbb P(A_i\cap A_j).
$$

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 Incoming links: Bonferroni inequalities