It is known that P(AB)=P(A)P(B) (i.e. the events A and B are…
It is known that $P(AB)=P(A)P(B)$ (i.e. the events $A$ and $B$ are independent), $C\supset AB$ and $\overline{C}\supset\overline{A}\,\overline{B}$. Prove that $P(AC)>P(A)P(C)$.
Explanation of Method
The conditions tell us exactly where $C$ can be: it contains the piece $AB$, it misses the piece $\overline{A}\,\overline{B}$, and it may contain parts of the two mixed pieces $A\overline{B}$ and $\overline{A}B$. We express $P(AC)$ and $P(C)$ through the probabilities of these parts, use independence to write $P(\overline{A}B)=P(B)(1-P(A))$, and compute the difference $P(AC)-P(A)P(C)$ exactly. It turns out to be a sum of two non-negative terms. So the inequality always holds with "$\ge$". It is strict in typical cases but can be an equality (for example when $C=B$), so the strict version must be stated with care.
Step 1: Translate the conditions
- $AB\subseteq C$: if both $A$ and $B$ occur, $C$ occurs.
- $\overline{A}\,\overline{B}\subseteq\overline{C}$: if neither $A$ nor $B$ occurs, $C$ does not occur. Equivalently (complements reverse inclusions) $C\subseteq A+B$.
- $P(AB)=P(A)P(B)$: $A$ and $B$ are independent.
Step 2: Split $\Omega$ into pieces and name the probabilities
Write $a=P(A)$, $b=P(B)$, so $P(AB)=ab$. Split $\Omega$ by $A$ and $B$ and look at how much of each piece lies in $C$:
| Piece | Part in $C$ | Part outside $C$ | Reason |
|---|---|---|---|
| $AB$ | all of it: probability $ab$ | $0$ | $AB\subseteq C$ |
| $A\overline{B}$ | $x=P(A\overline{B}C)$ | $P(A\overline{B}\,\overline{C})$ | no condition |
| $\overline{A}B$ | $y=P(\overline{A}BC)$ | $z=P(\overline{A}B\overline{C})$ | no condition |
| $\overline{A}\,\overline{B}$ | $0$ | all of it | $\overline{A}\,\overline{B}\subseteq\overline{C}$ |
Independence gives the probability of the piece $\overline{A}B$:
\[ P(\overline{A}B) = P(B) - P(AB) = b - ab = b(1-a), \qquad\text{so}\qquad y + z = b(1-a). \]
Step 3: $P(AC)$ and $P(C)$
$AC$ consists of the parts of $C$ inside $A$: the whole piece $AB$ and the part $A\overline{B}C$. So
\[ P(AC) = ab + x. \]
$C$ consists of $AB$, $A\overline{B}C$ and $\overline{A}BC$ (nothing from $\overline{A}\,\overline{B}$):
\[ P(C) = ab + x + y. \]
Step 4: Compute the difference
\[ \begin{aligned} P(AC) - P(A)P(C) &= ab + x - a(ab + x + y)\\ &= ab(1-a) + x(1-a) - ay. \end{aligned} \]
Now substitute $y = b(1-a) - z$ from Step 2:
\[ \begin{aligned} &= ab(1-a) + x(1-a) - ab(1-a) + az\\ &= (1-a)\,x + a\,z. \end{aligned} \]
So we get the exact formula
\[ \boxed{P(AC) - P(A)P(C) = P(\overline{A})\,P(A\overline{B}C) + P(A)\,P(\overline{A}B\overline{C}) \ \ge 0.} \]
Both terms are products of probabilities, so they are non-negative. Hence
\[ P(AC) \ge P(A)P(C). \]
Step 5: When is the inequality strict?
$P(AC) > P(A)P(C)$ holds exactly when at least one of the two terms is positive:
- $P(\overline{A})>0$ and $P(A\overline{B}C)>0$ ($C$ can occur together with $A$ but without $B$), or
- $P(A)>0$ and $P(\overline{A}B\overline{C})>0$ ($B$ can occur without $A$ and without $C$).
Equality can happen. Toss two fair coins; $A$ = "first coin heads", $B$ = "second coin heads" (independent), and take $C=B$. Then $AB\subseteq C$ and $\overline{A}\,\overline{B}\subseteq\overline{B}=\overline{C}$, so all conditions hold, but
\[ P(AC) = P(AB) = \tfrac14 = \tfrac12\cdot\tfrac12 = P(A)P(C). \]
Here both terms are zero: $A\overline{B}C=A\overline{B}B=\varnothing$ and $\overline{A}B\overline{C}=\overline{A}B\overline{B}=\varnothing$. So the strict inequality in the statement is true only under the extra condition above; in general the correct statement is $P(AC)\ge P(A)P(C)$.
A strict example. Same coins, $C = A+B$ ("at least one head"). Then $P(AC)=P(A)=\tfrac12$ and $P(A)P(C)=\tfrac12\cdot\tfrac34=\tfrac38$, so $P(AC)>P(A)P(C)$. The formula agrees: $P(\overline{A})P(A\overline{B}C) = \tfrac12\cdot\tfrac14 = \tfrac18 = \tfrac12-\tfrac38$. ✓
Step 6: Meaning
$P(AC)\ge P(A)P(C)$ means $A$ and $C$ are non-negatively correlated: knowing that $A$ occurred does not make $C$ less likely (if $P(A)>0$, it is equivalent to $P(C\mid A)\ge P(C)$). This is natural: $C$ is forced to occur when $A$ and $B$ both occur, and forced to fail when both fail, so $A$ "pushes" $C$ upwards.
Result
| Result | |
|---|---|
| Exact difference | $P(AC)-P(A)P(C) = P(\overline{A})P(A\overline{B}C) + P(A)P(\overline{A}B\overline{C})$ |
| Always true | $P(AC)\ \ge\ P(A)P(C)$ |
| Strict inequality | exactly when $P(\overline{A})P(A\overline{B}C)>0$ or $P(A)P(\overline{A}B\overline{C})>0$ |
| Equality example | $C=B$: then $P(AC)=P(A)P(C)$ |