Q.In a factory, machines $A$, $B$, $C$ manufacture $25\%$, $35\%$ and $40\%$ of the total output respectively. Of their outputs, $5\%$, $4\%$ and $2\%$ are defective respectively. An item is selected at random from the total output. Find the probability that it is defective.

[!TLDR] $P(D)=P(A)P(D\mid A)+P(B)P(D\mid B)+P(C)P(D\mid C)$. [!ANSWER] $P(D)=0.0345$ (i.e. $3.45\%$).

$P(D)=0.25\times0.05+0.35\times0.04+0.40\times0.02=0.0125+0.0140+0.0080=0.0345$. [!ANSWER] $P(\text{item is defective})=0.0345=3.45\%$.

Multiply each machine's output share by its own defective rate, then sum all three products (total probability theorem).

Averaging the three defective rates directly as $(5\%+4\%+2\%)/3$, ignoring the different output shares of each machine, gives a wrong, unweighted answer.