Venn Diagram vs Tree 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.

Each split in a tree is one stage of the process. Branches leaving the start carry unconditional probabilities, and every branch after that carries a probability conditional on the path so far, which is why multiplying along a path is the general multiplication rule $P(A \cap B) = P(A) \cdot P(B \mid A)$ drawn in picture form. Two mechanics run everything: multiply along a path, then add the paths that satisfy your event. Branches leaving any one split add to 1, and the complete paths add to 1 across the whole tree.

A plant runs two machines. Machine 1 makes 60 percent of the parts with a 2 percent defect rate; machine 2 makes the other 40 percent with a 5 percent defect rate. The tree has four paths. Machine 1 and defective is $0.60 \times 0.02 = 0.012$, machine 2 and defective is $0.40 \times 0.05 = 0.020$, so $P(\text{defective}) = 0.032$. The other two paths, 0.588 and 0.380, bring the total to 1.

The misreading is averaging the branch labels instead of weighting them: the rates are 2 percent and 5 percent, so overall it is 3.5 percent. That ignores that machine 1 turns out half again as many parts as machine 2, which pulls the true 3.2 percent below the midpoint. The plain average is right only when the stage-one branches are equal, or when the two rates were the same to begin with.

Reading a tree backwards takes division, not a branch label. Given that a part is defective, the chance it came from machine 2 is $0.020/0.032 = 0.625$, not the 0.05 printed on its branch.

A tree suits two or three stages. Paths double at every stage, so ten repeated trials would need 1,024 of them, and that is where the binomial formula takes over. Sampling without replacement does not break the method, it just changes the second-stage branches, which is the case worth drawing rather than reasoning through.