Probability of intersection of 3 events

If the Probabilities of 3 events are given i.e. $$\textbf{P}(A) = \textbf{P}(B) = \textbf{P}(C) \geq 0.99$$, then what is the minimum value of intersection of the 3 events?

• You might be able to figure this out for yourself by drawing a Venn diagram. – Gerry Myerson Sep 18 at 4:40

By inclusion–exclusion: $$1\ge P(A\cup B\cup C)=P(A)+P(B)+P(C)-(P(A\cap B)+P(B\cap C)+P(C\cap A))+P(A\cap B\cap C)\ge2.97-(P(A\cap B)+P(B\cap C)+P(C\cap A))+P(A\cap B\cap C)$$ which simplifies to $$1.97\le(P(A\cap B)+P(B\cap C)+P(C\cap A))-P(A\cap B\cap C)$$ Again, by inclusion–exclusion and the universal upper bound on all probabilities of $$1$$, $$P(A\cap B)+P(B\cap C)+P(C\cap A)\le1+2P(A\cap B\cap C)$$, so $$1.97\le1+2P(A\cap B\cap C)-P(A\cap B\cap C)=1+P(A\cap B\cap C)$$ $$0.97\le P(A\cap B\cap C)$$ This lower bound can be achieved by setting, other than $$0.97=P(A\cap B\cap C)$$, $$P(A\cap B\cap C^c)=P(C\cap A\cap B^c)=P(B\cap C\cap A^c)=0.01$$ and $$P(A\cap B^c\cap C^c)=P(C\cap A^c\cap B^c)=P(B\cap C^c\cap A^c)=0$$.
• @AnwayaRath The probability of the union of $A\cap B,B\cap C,C\cap A$ must, like all probabilities, be less than $1$. But we triple count $A\cap B\cap C$ if we blindly add $A\cap B,B\cap C,C\cap A$ together. Hence we add twice $P(A\cap B\cap C)$ to $1$ to obtain the upper bound. – Parcly Taxel Sep 18 at 7:21