Prove that if A_1A_2A_3⊂ A , then P(A) > P(A_1) + P(A_2) +…

Prove that if $A_1A_2A_3\subset A$, then \[ P(A) > P(A_1) + P(A_2) + P(A_3) - 2. \]

Step 1: Reduce to $P(A_1A_2A_3)$

The condition says: if $A_1$, $A_2$, $A_3$ all occur, then $A$ occurs. Since a smaller event has smaller or equal probability,

\[ P(A) \ge P(A_1A_2A_3). \tag{1} \]

Step 2: Pass to the opposite event

By De Morgan's law, "not all three occur" = "at least one of them fails":

\[ \overline{A_1A_2A_3} = \overline{A_1}+\overline{A_2}+\overline{A_3}, \qquad\text{so}\qquad P(A_1A_2A_3) = 1 - P(\overline{A_1}+\overline{A_2}+\overline{A_3}). \]

Step 3: Bound the probability of the sum (Boole's inequality)

The probability of a sum of events is at most the sum of their probabilities (overlaps are counted more than once on the right):

\[ P(\overline{A_1}+\overline{A_2}+\overline{A_3}) \le P(\overline{A_1})+P(\overline{A_2})+P(\overline{A_3}) = 3 - P(A_1) - P(A_2) - P(A_3). \]

Step 4: Combine

\[ P(A_1A_2A_3) \ge 1 - \big(3 - P(A_1) - P(A_2) - P(A_3)\big) = P(A_1)+P(A_2)+P(A_3) - 2. \]

Together with (1):

\[ \boxed{P(A) \ \ge\ P(A_1)+P(A_2)+P(A_3) - 2.} \]

Step 5: Second proof — use Problem 29 twice

From Problem 29 (Step 2), for any two events $P(XY)\ge P(X)+P(Y)-1$. Apply it with $X=A_1A_2$, $Y=A_3$, and then with $X=A_1$, $Y=A_2$:

\[ P(A_1A_2A_3) \ge P(A_1A_2) + P(A_3) - 1 \ge \big(P(A_1)+P(A_2)-1\big) + P(A_3) - 1 = P(A_1)+P(A_2)+P(A_3)-2. \ \checkmark \]

Step 6: Strict or not?

If $A_1=A_2=A_3=A=\Omega$, the condition holds and $P(A) = 1 = 1+1+1-2$: equality. So in general the inequality is "$\ge$". Equality needs both steps to be equalities: $P(A) = P(A_1A_2A_3)$ (Step 1), and no two of the "failures" $\overline{A_1},\overline{A_2},\overline{A_3}$ can happen together with positive probability (Step 3). In every other case the inequality is strict: $P(A) > P(A_1)+P(A_2)+P(A_3)-2$.

Step 7: Numerical example

A machine produces a good part ($A$) whenever the material is good ($A_1$), the tool is sharp ($A_2$) and the operator is attentive ($A_3$). If $P(A_1)=0.95$, $P(A_2)=0.9$, $P(A_3)=0.9$, then

\[ P(A) \ge 0.95 + 0.9 + 0.9 - 2 = 0.75, \]

whatever the dependence between the three conditions.

Step 8: General form for $n$ events

The same argument gives, for $A_1A_2\cdots A_n\subseteq A$,

\[ P(A) \ge P(A_1)+P(A_2)+\dots+P(A_n) - (n-1). \]

This is known as Bonferroni's inequality. Problem 29 is the case $n=2$.