Total probability theorem — solve using partition of sample space

Question

A factory has three machines $M_1$ , $M_2$ , and $M_3$ that produce 30%, 45%, and 25% of total output respectively. The defective rates for each machine are 2%, 3%, and 5%. A randomly selected item is found to be defective. Find the probability that it was produced by machine $M_2$ .

(JEE Main 2023, similar pattern — uses both total probability and Bayes’ theorem)

Solution — Step by Step

Let $E_i$ = event that the item comes from machine $M_i$ , and $D$ = event that item is defective.

$P(E_1) = 0.30$ , $P(E_2) = 0.45$ , $P(E_3) = 0.25$

$P(D|E_1) = 0.02$ , $P(D|E_2) = 0.03$ , $P(D|E_3) = 0.05$

The machines form a partition of the sample space (every item comes from exactly one machine). By the total probability theorem:

P(D) = P(E_1)P(D|E_1) + P(E_2)P(D|E_2) + P(E_3)P(D|E_3)

= 0.30 \times 0.02 + 0.45 \times 0.03 + 0.25 \times 0.05

= 0.006 + 0.0135 + 0.0125 = 0.032

P(E_2|D) = \frac{P(E_2)P(D|E_2)}{P(D)} = \frac{0.45 \times 0.03}{0.032} = \frac{0.0135}{0.032}

= \frac{135}{320} = \frac{27}{64} = \mathbf{0.421875}

The probability that the defective item came from $M_2$ is $\dfrac{27}{64} \approx 42.2\%$ .

Why This Works

The total probability theorem breaks a complex probability into simpler conditional pieces. We do not know $P(D)$ directly, but we know the defect rate for each machine. The theorem sums up the contributions from all sources.

Bayes’ theorem then “reverses” the conditioning: given that the item IS defective, which machine most likely produced it? Even though $M_3$ has the highest defect rate (5%), $M_2$ is more likely to be the source because it produces the most items (45% of output).

This “prior times likelihood” reasoning is the foundation of Bayesian statistics — a scoring topic in JEE.

Alternative Method

Work with actual numbers instead of probabilities. Assume 10,000 items produced:

$M_1$ : 3,000 items, 60 defective
$M_2$ : 4,500 items, 135 defective
$M_3$ : 2,500 items, 125 defective
Total defective: 320

$P(E_2|D) = 135/320 = 27/64$ . Same answer, and the counting approach is more intuitive.

For JEE and CBSE, Bayes’ theorem problems almost always follow this factory/machine template. Identify the “sources” (machines, bags, boxes), their proportions, and their conditional probabilities. The calculation is mechanical after that.

Common Mistake

Students confuse $P(D|E_2)$ with $P(E_2|D)$ . The first is the probability of a defect given it came from $M_2$ (= 0.03, given). The second is the probability it came from $M_2$ given it is defective (= 27/64, computed). These are very different numbers. Bayes’ theorem exists precisely to convert one into the other.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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