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 randomly selected. What is the probability that it is defective?​

Answer 1

Given :   three machines, B1, B2, and B3, make 30%, 45%, and 25%, respectively, of the products.

2%, 3%, and 2% of the products made by each machine, respectively, are defective.

a finished product is randomly selected.  

To Find :  What is the probability that it is defective

Solution:

Probability of Defective from machine B1  = (30/100)(2/100)  = 60/10000

Probability of  Defective from machine B2  = (45/100)(3/100)  = 135/10000

Probability of Defective from machine B3  = (25/100)(2/100)  = 50/10000

Probability of Defective  = (60 + 135 + 50)/10000

= 245/10000

= 0.0245

= 2.45 %  

0.0245   or  2.45 %    is the probability that it is defective

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Answer 2

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

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

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

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

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

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

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

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

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

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

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

$\mathsf\red{Hence}$

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

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

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

$\mathsf\red{So}$

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

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

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

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