A target consists of ten concentric circles with radii r_k…

A target consists of ten concentric circles with radii $r_k$ $(k = 1, 2, \dots, 10)$, where $r_1 \lt r_2 \lt \dots \lt r_{10}$. Event $A_k$ denotes a hit in the circle of radius $r_k$ $(k = 1, 2, \dots, 10)$. What do the events \[ B = \sum_{k=1}^{6} A_k \qquad\text{and}\qquad C = \prod_{k=1}^{10} A_k \] mean?

Explanation of Method

"A hit in the circle of radius $r_k$" means the hit lies inside that whole disc, so the events are nested: a smaller circle lies inside every larger one, hence $A_1\subseteq A_2\subseteq\dots\subseteq A_{10}$. For nested events the sum (union) is simply the largest event and the product (intersection) is simply the smallest event. Applying this to the first six events and to all ten events gives the answer.

Step 1: Understand the events $A_k$

Put the centre of the target at the origin and let $d$ be the distance from the hit to the centre. Then

\[ A_k = \{\text{the hit lies inside the circle of radius } r_k\} = \{d \le r_k\}. \]

Note that $A_k$ is the whole disc, not only the ring between two circles.

$C$ $B$ $r_6$ $r_{10}$

$B = A_6$: hit inside the circle of radius $r_6$
$C = A_1$: hit inside the innermost circle

Ten concentric circles. The discs are nested: a hit inside a smaller circle is also inside every larger one.

Step 2: The events are nested

If a hit satisfies $d\le r_k$, then, because $r_k \lt r_{k+1}$, it also satisfies $d \le r_{k+1}$. So every hit in $A_k$ is also in $A_{k+1}$:

\[ A_1 \subseteq A_2 \subseteq A_3 \subseteq \dots \subseteq A_{10}. \]

In words: the smaller the circle, the harder it is to hit; hitting a small circle automatically means hitting all the bigger ones.

Step 3: The event $B = A_1 + A_2 + \dots + A_6$

$B$ is the sum (union) of the first six events: it occurs if at least one of $A_1,\dots,A_6$ occurs. We show $B = A_6$ by proving two inclusions.

$B\subseteq A_6$: each of $A_1,\dots,A_6$ lies inside $A_6$ (Step 2). If the hit is in one of them, it is in $A_6$.
$A_6\subseteq B$: $A_6$ is itself one of the terms of the sum, so a hit in $A_6$ makes $B$ occur.

\[ \boxed{B = \sum_{k=1}^{6} A_k = A_6}. \]

Meaning: $B$ is a hit inside the circle of radius $r_6$ (anywhere within the sixth circle, including the five smaller circles inside it), i.e. at distance at most $r_6$ from the centre.

Step 4: The event $C = A_1 A_2 \cdots A_{10}$

$C$ is the product (intersection) of all ten events: it occurs if all of $A_1,\dots,A_{10}$ occur at the same time. We show $C = A_1$.

$C\subseteq A_1$: if all ten events occur, then in particular $A_1$ occurs.
$A_1\subseteq C$: if the hit is in $A_1$, then by Step 2 it is in $A_2, A_3, \dots, A_{10}$ as well, so all ten events occur.

\[ \boxed{C = \prod_{k=1}^{10} A_k = A_1}. \]

Meaning: $C$ is a hit inside the smallest (innermost) circle of radius $r_1$ — the bullseye. It is the only region that lies inside all ten circles.

Step 5: The general rule

For any nested events $S_1\subseteq S_2\subseteq\dots\subseteq S_n$:

\[ S_1 + S_2 + \dots + S_n = S_n \ \ (\text{the largest}), \qquad S_1S_2\cdots S_n = S_1 \ \ (\text{the smallest}). \]

A sum can never be smaller than its largest term, and a product can never be larger than its smallest factor; for nested events these bounds are reached.

Step 6: A remark on a different reading

If $A_k$ meant "a hit in the $k$-th ring" (between circles $r_{k-1}$ and $r_k$), the events would be mutually exclusive. Then $B$ would again be "a hit inside the circle of radius $r_6$", but $C$ would be the impossible event $\varnothing$, since one hit cannot land in ten different rings. The problem says "a hit in the circle", so the nested reading above is the correct one.

Result

Event	Simplifies to	Meaning
$B=\sum_{k=1}^{6}A_k$	$A_6$	a hit inside the circle of radius $r_6$
$C=\prod_{k=1}^{10}A_k$	$A_1$	a hit inside the innermost circle of radius $r_1$ (bullseye)