A factory has two Machines - $I$ and $II$. Machine-$I$ produces $25\%$ of items and Machine-$II$ produces $75\%$ of the items of the total output. Further $3\%$ of the items produced by Machine-$I$ are defective whereas $4\%$ produced by Machine-$II$ are defective. If an item is drawn at random, what is the probability that it is defective?
Let $A_1$ be the event that the items are produced by machine $I$
Let $A_2$ be the event that the items are produced by machine $II$
Let $B$ be the event of selecting a defective item
To find: $\;$ P(of selecting a defective item) $= P \left(B\right)$
Events $A_1$ and $A_2$ are mutually exclusive and exhaustive events.
$\therefore \;$ $P \left(B\right) = P \left(A_1\right) \times P \left(B | A_1\right) + P \left(A_2\right) \times P \left(B | A_2\right)$ $\;\;\; \cdots \; (1)$
Probability that the items are produced by machine-$I$ $= P \left(A_1\right) = \dfrac{25}{100}$
Probability that the items are produced by machine-$II$ $= P \left(A_2\right) = \dfrac{75}{100}$
P(getting a defective item when the item is selected from machine-$I$)
$= P \left(B | A_1\right) = \dfrac{3}{100}$
P(getting a defective item when the item is selected from machine-$II$)
$= P \left(B | A_2\right) = \dfrac{4}{100}$
Substituting the values of $P \left(A_1\right)$, $P \left(A_2\right)$, $P \left(B | A_1\right)$ and $P \left(B | A_2\right)$ in equation $(1)$ we have,
$P \left(B\right) = \left(\dfrac{25}{100} \times \dfrac{3}{100}\right) + \left(\dfrac{75}{100} \times \dfrac{4}{100}\right) = \dfrac{3}{80}$