Answer
$0.027$
Work Step by Step
Define the events
$A_{1}$ = "Bearing came from supplier $1$."
$A_{2}$ = "Bearing came from supplier $2$."
$A_{3}$ = "Bearing came from supplier $3$."
$B$ = "Bearing is defective"
We are given:
$P(A_{1})=0.5$
$P(A_{2})=0.3$
$P(A_{3})=0.2$
$P(B|A_{1})=0.02, P(B|A_{2})=0.03, P(B|A_{3})=0.04$
Now apply the theorem:
$$\begin{align*}
P(B)&=\displaystyle \sum_{i=1}^{n}P(B|A_{i})P(A_{i}) & & \\
&=(0.02)(0.5)+(0.03)(0.3)+(0.04)(0.2) \\
& =0.027 \end{align*}$$
