Probability: Application Problems (3)

Question

In a factory, machines $A$, $B$ and $C$ produce $25\%$, $35\%$ and $40\%$ of the total bolts respectively. Of the bolts produced, $5\%$, $4\%$ and $2\%$ are defective from $A$, $B$ and $C$ respectively. A bolt is drawn at random and found defective. What is the probability it was made by machine $A$?

Solution — Step by Step

Why This Works

Bayes’ theorem flips conditional probabilities. We knew $P(\text{defect} | \text{machine } A)$, but the question asks $P(\text{machine } A | \text{defect})$. The theorem gives us the reverse using the law of total probability for the denominator.

A useful sanity check: $P(A|D) + P(B|D) + P(C|D)$ must equal $1$. Computing $P(B|D) = 0.014/0.0345 = 28/69$ and $P(C|D) = 16/69$, we get $25 + 28 + 16 = 69$ ✓.

Alternative Method

Imagine $1000$ bolts. Machine $A$ produces $250$, of which $5\%$ ($12.5$) are defective. Machine $B$: $350 \times 4\% = 14$ defective. Machine $C$: $400 \times 2\% = 8$ defective. Total defective: $34.5$. Of these, machine A made $12.5$, so $P(A|D) = 12.5/34.5 = 25/69$. Same answer, no theorem needed.

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