Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 2 / 11F (Probability)

The statement is true. Put $C_k={\bigcap }_{r=1}^k{A}_r$. The assumed $P(C_{n-1})>0$ ensures that every conditioning event $C_{k-1}$, $2\le k\le n$, has positive probability. By the definition of conditional probability,Multiplying makes the fractions telescope, givingThis proves the chain rule for probabilities here without any assumption of independence. The final numerator $P(C_n)$ may be zero; all denominators remain positive.

The statement is true. Use the union bound on the complementary events:This proves the claim for arbitrary events, with no independence requirement.

The statement is true. For $n\ge 2$, being in every pairwise intersection is equivalent to being in all the events:Indeed every index occurs in at least one pair. Apply the union bound to the complements of these $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ pairwise intersections to obtainThis is positive common intersection from pair-intersection probabilities. It remains valid for $n=2$, where it reduces to the asserted positivity of the single pairwise intersection.

The statement is false. Take two independent fair bits $U,V$ on the four equally probable outcomes $00,01,10,11$, and put $A_1=\{U=0\}$, $A_2=\{V=0\}$, and $B=\{U=V\}$. Each of $A_1,A_2,B$ has probability $1/2$, andThus $B$ is independent of each event separately. But $A_1\cup A_2$ contains $00,01,10$, whereas its intersection with $B$ contains only $00$. ThereforeThis is a concrete failure of extending separate independence to the generated union. A counterexample at the permitted value $n=2$ disproves the universal claim; for any larger $n$ one can repeat $A_1$ as the additional events.