Question

In a certain assembly Plant, three machines B1,B2, B3 produce 30%,45% A
25% of the products, respectively. It is known from the past experience that
2%,3% 12% of the product gade by each machine, respectively, are
defective. Supposed that a finished product is randomly selected. What is the
probability that it is defective?​

Answer 1

Answer:

HEY LOOK THROUGH IT !

Step-by-step explanation:

P

A

=

P

B1 P

A

B1 +

P

B2

P

A

B2

+

P

B3

P

A

B3

It is clear that:

(

)

(

)

(

0.3)(

0.02)

0.006, P

B1 P

A

B1 =

=

(

)

(

)

(

0.45)(

0.03)

0.0135

P

B2

P

A

B2

=

=

(

)

(

)

(

0.25)(

0.02)

0.005

P

B3

P

A

B3

=

= Hence

P

(

A

)

=

0.006

+

0.0135

+

0.005

=

0.0245

Answer 2

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

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

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

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

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

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

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

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

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

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

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

$\mathsf\red{Hence}$

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

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

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

$\mathsf\red{So}$

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

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

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

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