Question

In a certain assembly plant, three machines B1, B2, and B3, make 30%, 45%, and 25%, respectively, of the products. It is known from past experience that 2%, 3% and 2% of the products made by each machine, respectively, are defective. Now, suppose that a finished product is random selected. What is the probability that it is defective?​

Answer 1

$\mathsf\red{Consider \: the \: following \: events \: :}$

$\mathsf{A \: : \: The \: product \: is \: defective}$

$\mathsf{B_{1} \: : \: The \: product \: is \: made \: by \: machine \: B_{1}}$

$\mathsf{B_{2} \: : \: The \: product \: is \: made \: by \: machine \: B_{2}}$

$\mathsf{B_{3} \: : \: The \: product \: is \: made \: by \: machine \: B_{3}}$

$\mathsf\red{Using \: the \: theorem \: of \: total \: probability \: :}$

$\mathsf{P(A) \: = \: P(B_{1}).P(A|B_{1})+P(B_{2}).P(A|B_{2})+P(B_{3}).P(A|B_{3})}$

$\mathsf\red{But \: We \: have \: :}$

$\mathsf{P(B_{1}) \: = \: 0.30 , \: P(A|B_{1}) \: = \: 0.02}$

$\mathsf{P(B_{2}) \: = \: 0.45 , \: P(A|B_{2}) \: = \: 0.03}$

$\mathsf{P(B_{3}) \: = \: 0.25 , \: P(A|B_{3}) \: = \: 0.02}$

$\mathsf\red{Hence}$

$\mathsf{P(B_{1}).P(A|B_{1}) \: = \: 0.3×0.02 \: = \: 0.006}$

$\mathsf{P(B_{2}).P(A|B_{2}) \: = \: 0.45×0.03 \: = \: 0.0135}$

$\mathsf{P(B_{3}).P(A|B_{3}) \: = \: 0.25×0.02 \: = \: 0.005}$

$\mathsf\red{So}$

$\mathsf{P(A) \: = \: P(B_{1}).P(A|B_{1})+P(B_{2}).P(A|B_{2})+P(B_{3}).P(A|B_{3})}$

$\mathsf{P(A) \: = \: 0.006 + 0.0135 + 0.005}$

$\mathsf{P(A) \: = \: 0.0245}$

$\mathsf{}$

$\large{\sf{\fcolorbox{black}{pink}{\copyright \: \: \: ITZBFF}}}$

Answer 2

$\mathsf\red{Consider \: the \: following \: events \: :}$

$\mathsf{A \: : \: The \: product \: is \: defective}$

$\mathsf{B_{1} \: : \: The \: product \: is \: made \: by \: machine \: B_{1}}</p><p>$

$</p><p>\mathsf{B_{2} \: : \: The \: product \: is \: made \: by \: machine \: B_{2}}$

$\mathsf{B_{3} \: : \: The \: product \: is \: made \: by \: machine \: B_{3}}</p><p>$

$\mathsf\red{Using \: the \: theorem \: of \: total \: probability \: :}$

$\mathsf{P(A) \: = \: P(B_{1}).P(A|B_{1})+P(B_{2}).P(A|B_{2})+P(B_{3}).P(A|B_{3})}$

$\mathsf\red{But \: We \: have \: :}$

$\mathsf{P(B_{1}) \: = \: 0.30 , \: P(A|B_{1}) \: = \: 0.02}$

$\mathsf{P(B_{2}) \: = \: 0.45 , \: P(A|B_{2}) \: = \: 0.03}$

$\mathsf{P(B_{3}) \: = \: 0.25 , \: P(A|B_{3}) \: = \: 0.02}$

$\mathsf\red{Hence}$

$\mathsf{P(B_{1}).P(A|B_{1}) \: = \: 0.3×0.02 \: = \: 0.006}$

$\mathsf{P(B_{2}).P(A|B_{2}) \: = \: 0.45×0.03 \: = \: 0.0135}$

$\mathsf{P(B_{3}).P(A|B_{3}) \: = \: 0.25×0.02 \: = \: 0.005}$

$\mathsf\red{So}$

$\mathsf{P(A) \: = \: P(B_{1}).P(A|B_{1})+P(B_{2}).P(A|B_{2})+P(B_{3}).P(A|B_{3})}$

$\mathsf{P(A) \: = \: 0.006 + 0.0135 + 0.005}$

$\mathsf{P(A) \: = \: 0.0245}$

$\mathsf\red{\therefore \: 0.0245 \: is \: the \: probability \: of \: defective}$