Tree diagram

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.