Events One of Which equals Intersection

Theorem

Let the probability space of an experiment $\EE$ be $\struct {\Omega, \Sigma, \Pr}$.

Let $A, B \in \Sigma$ be events of $\EE$, so that $A \subseteq \Omega$ and $B \subseteq \Omega$.

Let $A$ and $B$ be such that:

$A \cap B = A$

Then whenever $A$ occurs, it is always the case that $B$ occurs as well.

$A \cap B = A \iff A \subseteq B$

Let $A$ occur.

Let $\omega$ be the outcome of $\EE$.

Let $\omega \in A$.

That is, by definition of occurrence of event, $A$ occurs.

Then by definition of subset:

$\omega \in B$

Thus by definition of occurrence of event, $B$ occurs.

Hence the result.

$\blacksquare$

Let $T$ be a target which consists of $10$ concentric circles $C_1$ to $C_{10}$ whose radii are respectively $r_k$ for $k = 1, 2, \ldots, 10$.

Let $r_k < r_{k + 1}$ for all $k = 1, 2, \ldots, 9$.

That is, let $C_1$ be the innermost and $C_{10}$ be the outermost.

Let $A_k$ denote the event of hitting $T$ inside the circle of radius $r_k$.

Let $C$ denote the event:

$C = \ds \bigcap_{k \mathop = 5}^{10} A_k$

Then $C$ is the event of hitting $T$ inside circle $C_5$.