Probability axioms and theorems — addition, multiplication, Bayes' theorem when to use

Question

A factory has two machines. Machine A produces 60% of items and Machine B produces 40%. Defective rates are 2% for A and 5% for B. If a randomly selected item is found defective, what is the probability it came from Machine A?

(Bayes’ theorem — CBSE 12 Board favourite, JEE Main 2022 pattern)

Which Theorem to Use?

flowchart TD
    A["Probability Problem"] --> B{What type?}
    B -->|"P(A or B)"| C["Addition Theorem"]
    B -->|"P(A and B)"| D["Multiplication Theorem"]
    B -->|"Given outcome, find cause"| E["Bayes' Theorem"]
    C --> C1{"Mutually exclusive?"}
    C1 -->|Yes| C2["P(A or B) = P(A) + P(B)"]
    C1 -->|No| C3["P(A or B) = P(A) + P(B) - P(A and B)"]
    D --> D1{"Independent?"}
    D1 -->|Yes| D2["P(A and B) = P(A) x P(B)"]
    D1 -->|No| D3["P(A and B) = P(A) x P(B|A)"]
    E --> E1["P(cause|effect) = P(effect|cause) x P(cause) / P(effect)"]

Solution — Step by Step

We observe the effect (item is defective) and want the cause (which machine made it). Whenever the problem gives you conditional probabilities and asks you to “reverse” the conditioning, that is Bayes’ theorem.

Let $A$ = item from Machine A, $B$ = item from Machine B, $D$ = item is defective.

$P(A) = 0.6$ , $P(B) = 0.4$
$P(D|A) = 0.02$ , $P(D|B) = 0.05$

P(D) = P(D|A) \cdot P(A) + P(D|B) \cdot P(B)

P(D) = 0.02 \times 0.6 + 0.05 \times 0.4 = 0.012 + 0.020 = 0.032

P(A|D) = \frac{P(D|A) \cdot P(A)}{P(D)} = \frac{0.02 \times 0.6}{0.032} = \frac{0.012}{0.032}

\boxed{P(A|D) = \frac{3}{8} = 0.375}

Even though Machine A produces more items, it has a lower defective rate. So only 37.5% of defective items come from A.

Why This Works

Bayes’ theorem is just conditional probability written backwards. We know $P(D|A)$ (probability of defect given Machine A), but we want $P(A|D)$ (probability of Machine A given defect). The formula connects the two using the total probability in the denominator.

The denominator $P(D)$ acts as a normalising constant — it ensures the posterior probabilities $P(A|D) + P(B|D) = 1$ .

Alternative Method — Tree Diagram Approach

Draw a probability tree: first branch into A (0.6) and B (0.4), then each branches into Defective and Not Defective.

Branch A-Defective: $0.6 \times 0.02 = 0.012$
Branch B-Defective: $0.4 \times 0.05 = 0.020$

$P(A|D) = \dfrac{0.012}{0.012 + 0.020} = \dfrac{0.012}{0.032} = \dfrac{3}{8}$

In CBSE boards, Bayes’ theorem appears every single year — usually as a 4-6 mark question. The setup is always the same: two or three “causes” with known conditional probabilities, and you reverse the conditioning. Practice 10 such problems and the pattern becomes automatic.

Common Mistake

The most common error: confusing $P(D|A)$ with $P(A|D)$ . Students write $P(A|D) = 0.02$ directly, thinking “defect rate of A” answers the question. But 0.02 is the probability that an item from A is defective, not the probability that a defective item came from A. These are fundamentally different — Bayes’ theorem bridges the gap.

Another mistake: forgetting to include all causes in the denominator. If there were three machines, $P(D) = P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C)$ .