Q.A factory has two machines A and B. Past record shows that machine A produced $60\%$ of the items of output and machine B produced $40\%$ of the items. Further, $2\%$ of the items produced by machine A and $1\%$ produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B?

Concept: Conditional Probability (Bayes' Theorem)

Let $D$ be the event that an item is defective.

We are given:

$P(A) = 0.6$, $P(B) = 0.4$,

$P(D|A) = 0.02$, $P(D|B) = 0.01$.

We need $P(B|D)$.

Step 1: Find $P(D)$ using the law of total probability:

$P(D) = P(A)P(D|A) + P(B)P(D|B) = (0.6)(0.02) + (0.4)(0.01) = 0.012 + 0.004 = 0.016$.

Step 2: Apply Bayes' Theorem: …

We are dealing with a classic inverse probability problem. The question gives us the overall production shares (the priors) and the defect rates within each machine (the likelihoods). We observe a defective item and want the chance it came from machine B — that is, we need to reverse the conditional probability.

The tool for this is Bayes’ theorem, which in its simplest form says:

where $D$ is the event “item is defective”. The denominator $P(D)$ is the total probability of a defective item, found by weighting each machine’s defect rate by its production share.

Let’s assign events clearly:

• Let $A$ = item produced by machine A.
• Let $B$ = item produced by machine B.
• Let $D$ = item is defective.

From the problem:

• $P(A) = 0.60$, $P(B) = 0.40$
• $P(D \mid A) = 0.02$ (2% of A’s items are defective)
• $P(D \mid B) = 0.01$ (1% of B’s items are defective)

We want $P(B \mid D)$.

1. Find the total probability of a defective item, $P(D)$. By the law of total probability:

Substitute:

So 1.6% of all items in the stockpile are defective.

2. Apply Bayes’ theorem:

Compute the numerator: $0.40 \times 0.01 = 0.004$.

Then:

3. Interpret the result: …

Method: Bayes' Theorem — tracing a defect back to its source

Use this whenever items from several producers are pooled, one is inspected, and you must decide which producer it most likely came from.

Steps

Step 1: Record each source's production share as its prior.

$P(H_i)$ is the fraction of all items made by source $i$.

Step 2: Record each source's defect (evidence) rate.

$P(E \mid H_i)$ is the chance an item from source $i$ shows the observed feature (defective).

Step 3: Total probability of the evidence across the pooled stockpile.

$P(E) = \sum_i P(H_i)\,P(E \mid H_i).$ …

Common Mistakes

Mistake 1: Comparing only the defect rates ($1\%$ vs $2\%$) and answering $\tfrac13$.

Why it's wrong: it ignores that A makes $60\%$ of the items and B only $40\%$. Correct approach: weight each defect rate by its production share via Bayes' theorem.

Mistake 2: Leaving out machine A's contribution to $P(\text{defective})$.

Why it's wrong: the defective item could have come from either machine, so both terms are needed. Correct approach: $P(D)=(0.6)(0.02)+(0.4)(0.01)=0.016$. …