Venn diagram

A Venn diagram draws the sample space as a rectangle and each event as a circle inside it. Two events cut it into four non-overlapping regions that cover everything: $A$ only, the lens where both happen ($A \cap B$), $B$ only, and the outside area, which is $(A \cup B)^c$. Every outcome sits in exactly one region. A complement is taken against the rectangle, not the other circle, so $A^c$ holds the $B$-only region and the outside.

In a class of 30, 18 play a sport, 12 play an instrument, and 7 do both. Put the 7 in the lens first. Sport only is $18 - 7 = 11$, instrument only is $12 - 7 = 5$, and the outside region is $30 - (11 + 7 + 5) = 7$. A student drawn at random does at least one with probability $23/30 \approx 0.7667$, matching the addition rule value $\frac{18}{30} + \frac{12}{30} - \frac{7}{30}$.

"18 play a sport and 12 play an instrument, so write 18 in the left circle and 12 in the right." Those totals describe whole circles, not the crescent regions. Filled in that way the diagram holds $18 + 7 + 12 = 37$ students in a class of 30, because the 7 who do both were entered twice.

Do not check the work by adding the four regions to 30. If the outside region came from subtracting the other three, that sum is forced and cannot fail. Two checks can. No region may be negative, so had the problem said 14 do both, instrument only would be $12 - 14 = -2$ and the counts impossible. And when the problem states a count the diagram also produces, such as the 7 who do neither, compare them: agreement there is real evidence, since the two came from different inputs.

The circles carry no scale, so read the numbers in the regions, never the areas. Mutually exclusive events are circles that do not touch. Three events push the picture to eight regions, where a tree diagram is easier to keep straight.