A square D contains two regions D_1 and D_2 with areas S_1…

A square $D$ contains two regions $D_1$ and $D_2$ with areas $S_1$ and $S_2$. A point is thrown into $D$ and can fall into any part of the square with equal probability. Let event $A_k$ denote that the point falls into region $D_k$ $(k = 1, 2)$. Are the events $A_1$ and $A_2$ equally likely?

Step 1: Geometric probability

Let $S$ be the area of the square $D$. When a point is thrown "uniformly" into $D$, regions of the same area have the same chance of being hit, and a region twice as large has twice the chance. This is expressed by the formula

\[ P(\text{point falls into } G) = \frac{\text{area of } G}{\text{area of } D}, \qquad G\subseteq D. \]

Step 2: The probabilities of $A_1$ and $A_2$

\[ P(A_1) = \frac{S_1}{S}, \qquad P(A_2) = \frac{S_2}{S}. \]

Step 3: When are they equally likely?

Two events are equally likely when they have the same probability. Since $S > 0$ we may multiply by $S$:

\[ P(A_1) = P(A_2) \iff \frac{S_1}{S} = \frac{S_2}{S} \iff S_1 = S_2. \]

So the answer is:

\[ \boxed{A_1 \text{ and } A_2 \text{ are equally likely if and only if } S_1 = S_2.} \]

If $S_1 \gt S_2$, the event $A_1$ is more likely; if $S_1 \lt S_2$, the event $A_2$ is more likely. Without knowing the areas we cannot say that the events are equally likely.

Step 4: Shape and position do not matter

Only the area enters the formula. A disc and a square, a long thin strip and a triangle — if they have the same area, a random point is equally likely to land in either of them. It also does not matter whether $D_1$ and $D_2$ overlap: the formula for each probability uses only that region's own area. (Overlap only means that both events can occur at once, i.e. $A_1A_2\neq\varnothing$.)

Step 5: Examples

Example 1: equal areas — equally likely

Let $D$ be a square of side $10$, so $S = 100$. Let $D_1$ be a disc of radius $2$ and $D_2$ a square of side $2\sqrt{\pi}$ (see the figure). Then

\[ S_1 = \pi\cdot 2^2 = 4\pi, \qquad S_2 = \left(2\sqrt{\pi}\right)^2 = 4\pi, \qquad P(A_1) = P(A_2) = \frac{4\pi}{100} \approx 0.126. \]

The events are equally likely, although the regions look completely different.

Example 2: different areas — not equally likely

Let $D$ be the unit square ($S = 1$), $D_1$ its left half ($S_1 = 0.5$) and $D_2$ its top quarter ($S_2 = 0.25$). Then $P(A_1) = 0.5 \ne 0.25 = P(A_2)$: the point is twice as likely to land in $D_1$.

Example 3: one region inside the other

If $D_1\subseteq D_2$ and $D_2$ has strictly larger area, then $A_1 \subseteq A_2$ and $P(A_1) \lt P(A_2)$ — not equally likely.

Step 6: "Possible" versus "equally likely"

Do not confuse two different ideas:

• An event is possible if it can happen at all. Here both events are possible as soon as $S_1 \gt 0$ and $S_2 \gt 0$.
• Two events are equally likely if they have the same probability. Here this requires $S_1 = S_2$.