The event A is a special case of the event B (that is, A⊂eq…
The event $A$ is a special case of the event $B$ (that is, $A\subseteq B$: whenever $A$ occurs, $B$ occurs too). Prove that $P(A)\le P(B)$.
Explanation of Method
Split $B$ into the part that lies in $A$ (which is all of $A$) and the part outside $A$. These two parts are mutually exclusive, so $P(B)$ is the sum of their probabilities. Since probabilities are non-negative, $P(B)$ is at least $P(A)$.
Step 1: Split $B$
Since $A\subseteq B$, we have $AB=A$ and
\[ B=AB+\overline AB=A+\overline AB, \qquad A\cdot\overline AB=\varnothing. \]
Step 2: Addition rule
\[ P(B)=P(A)+P(\overline AB). \]
Step 3: Non-negativity
$P(\overline AB)\ge0$, so
\[ \boxed{P(A)\le P(B).} \quad\blacksquare \]
Moreover, the difference is exactly $P(B)-P(A)=P(\overline AB)$, the probability that $B$ occurs without $A$.
Step 4: Example
A die: $A$ = "a six", $B$ = "an even number". A six is a special case of an even number, and $\frac16\le\frac12$. The difference $\frac13$ is the probability of $\{2,4\}$.
Note: $P(A)=P(B)$ does not imply $A=B$; it only means that $P(\overline AB)=0$.
Result
| Decomposition | $B=A+\overline AB$ (incompatible) |
| Conclusion | $P(B)=P(A)+P(\overline AB)\ge P(A)$ |