Question

A factory production line is manufacturing bolts using three machines, A,B and C. Of the total output, machine A is responsible for 20%, machine B for 30% and machine C for the rest. It is known from previous experience with the machines that 3% of the output from machine A is defective, 4% from machine B and 2% from machine C. A bolt is chosen at random from the production line.

(1)What is the probability that it is defective?

(2)If we learn that the bolt selected is defective, what is the probability that it came from machine B?​

Answer 1

$\large\underline{\sf{Solution-}}$

Let us suppose that

$\sf \:\: E_1 : \: bolt \: is \: manufactured \: by \: machine \: A$

$\sf \:\: E_2 : \: bolt \: is \: manufactured \: by \: machine \: B$

$\sf \:\: E_3 : \: bolt \: is \: manufactured \: by \: machine \: C$

$\sf \: E : getting \: a \: defective \: bolt$

$\sf \: E_1, \: E_2, \: E_3 \: are \: mutually \: exclusive \: and \: exhaustive \: events.$

According to statement,

$\sf \: \: \: \: \: \: \: \: \: \bull \: \: P(E_1) = \dfrac{20}{100}$

$\sf \: \: \: \: \: \: \: \: \: \bull \: \: P(E_2) = \dfrac{30}{100}$

$\sf \: \: \: \: \: \: \: \: \: \bull \: \: P(E_3) = \dfrac{50}{100}$

Now,

Probability of getting defective bolt manufactured by machine A is

$\sf \: \: \: \: \: \: \: \: \: \bull \: \: P(E | E_1) = \dfrac{3}{100}$

Probability of getting defective bolt manufactured by machine B is

$\sf \: \: \: \: \: \: \: \: \: \bull \: \: P(E | E_2) = \dfrac{4}{100}$

Probability of getting defective bolt manufactured by machine C is

$\sf \: \: \: \: \: \: \: \: \: \bull \: \: P(E | E_3) = \dfrac{2}{100}$

Now,

Probability of getting defective bolt is

$\sf \: P(E) = P(E_1)P(E | E_1) + P(E_2)P(E | E_2) + P(E_3)P(E | E_3)$

$\sf \: P(E) = \dfrac{20}{100} \times \dfrac{3}{100} + \dfrac{30}{100} \times \dfrac{4}{100} + \dfrac{50}{100} \times \dfrac{2}{100}$

$\sf \: P(E) = \dfrac{280}{10000} = 0.028$

Now, the required probability that bolt is defective and manufactured by machine B,

Using Baye's Theorem, is given by

$\sf \: P(E_2 | E) = \dfrac{P(E_2)P(E | E_2)}{P(E_1)P(E | E_1) + P(E_2)P(E | E_2) + P(E_3)P(E | E_3)}$

$\sf \: \: \: = \: \: \dfrac{\dfrac{30}{100} \times \dfrac{4}{100} }{\dfrac{20}{100} \times \dfrac{3}{100} + \dfrac{30}{100} \times \dfrac{4}{100} + \dfrac{50}{100} \times \dfrac{2}{100} }$

$\sf \: \: \: = \: \dfrac{\dfrac{120}{10000} }{\dfrac{280}{10000} }$

$\sf \: \: \: = \: \: \dfrac{3}{7}$