It is known that the simultaneous occurrence of the events…

It is known that the simultaneous occurrence of the events $A_1$ and $A_2$ necessarily implies the occurrence of the event $A$. Prove that \[ P(A) > P(A_1) + P(A_2) - 1. \]

Step 1: Translate the condition

"Whenever $A_1$ and $A_2$ both occur, $A$ occurs" means

\[ A_1A_2 \subseteq A. \]

A smaller event cannot have larger probability (monotonicity: if $X\subseteq Y$ then $Y = X + \overline{X}Y$, so $P(Y) = P(X) + P(\overline{X}Y)\ge P(X)$). Hence

\[ P(A) \ge P(A_1A_2). \tag{1} \]

Step 2: A lower bound for $P(A_1A_2)$

By the addition rule,

\[ P(A_1+A_2) = P(A_1) + P(A_2) - P(A_1A_2). \]

Every probability is at most $1$, so $P(A_1)+P(A_2)-P(A_1A_2)\le 1$, i.e.

\[ P(A_1A_2) \ge P(A_1) + P(A_2) - 1. \tag{2} \]

In words: if two events are both very likely, they must overlap a lot.

Step 3: Combine

From (1) and (2):

\[ \boxed{P(A) \ \ge\ P(A_1A_2) \ \ge\ P(A_1) + P(A_2) - 1.} \]

Step 4: Can the inequality be strict ("$>$")?

Not always. Take $A_1 = A_2 = A = \Omega$ (the certain event). The condition $A_1A_2\subseteq A$ holds, and

\[ P(A) = 1 = 1 + 1 - 1 = P(A_1)+P(A_2)-1. \]

So equality is possible, and the correct statement is with "$\ge$". Looking at Steps 1–2, equality holds exactly when both inequalities are equalities:

• $P(A) = P(A_1A_2)$: $A$ has no extra probability outside $A_1A_2$, and
• $P(A_1+A_2) = 1$: at least one of $A_1$, $A_2$ occurs with probability $1$.

In every other case the inequality is strict: $P(A) > P(A_1)+P(A_2)-1$.

Step 5: Numerical example

A component works ($A$) if its power supply works ($A_1$) and its controller works ($A_2$) — and possibly also in other situations. If $P(A_1)=0.9$ and $P(A_2)=0.8$, then without knowing anything else,

\[ P(A) \ge 0.9 + 0.8 - 1 = 0.7. \]

Note: if $P(A_1)+P(A_2)\le 1$, the bound is $\le 0$ and tells us nothing new (probabilities are always $\ge 0$).

Step 6: Another proof via opposite events

$\overline{A_1A_2} = \overline{A_1}+\overline{A_2}$ (De Morgan), and the probability of a sum is at most the sum of probabilities:

\[ P(A_1A_2) = 1 - P(\overline{A_1}+\overline{A_2}) \ge 1 - P(\overline{A_1}) - P(\overline{A_2}) = P(A_1)+P(A_2)-1. \]

This form generalises easily to three or more events (see Problem 30).