Question

Three companies A, B and C supply 25%, 35% and 40% of the notebooks to a school. Past experience shows that 5%, 4% and 2% of the notebooks produced by these companies are defective. If a notebook was found to be defective, what is the probability that the notebook was supplied by A

Answer 1

Answer:b

Step-by-step explanation:

Let A, B and C be the events that notebooks are provided by A, B and C respectively.

Let D be the event that notebooks are defective

Then,

P(A) = 0.25, P(B) = 0.35, P(C) = 0.4

P(D|A) = 0.05, P(D|B) = 0.04, P(D|C) = 0.02

P(A│D) = (P(D│A)*P(A))/(P(D│A) * P(A) + P(D│B) * P(B) + P(D│C) * P(C) )

= (0.05*0.25)/((0.05*0.25)+(0.04*0.35)+(0.02*0.4)) = 2000/(80*69)

= 25⁄69.

Answer 2

Given: Three companies A, B and C supply 25%, 35% and 40% of the notebooks to a school. Past experience shows that 5%, 4% and 2% of the notebooks produced by these companies are defective.

To find: If a notebook was found to be defective, what is the probability that the notebook was supplied by A.

Solution: Let E1= Notebook was supplied by A

Since A supplies 25% of the notebook,

P(E1) = 25/100 = 0.25

Let E2= Notebook was supplied by B

Since A supplies 35% of the notebook,

P(E2) = 35/100 = 0.35

Let E3= Notebook was supplied by C

Since A supplies 40% of the notebook,

P(E3) = 40/100 = 0.4

Let P(A) be the probability that the notebook was defective.

Then, P(A/E1) = Notebook was defective when it was supplied by A

= 5% of the total notebooks

= 5/100

= 0.05

P(A/E2) = Notebook was defective when it was supplied by B

= 4% of the total notebooks

= 4/100

= 0.04

P(A/E3) = Notebook was defective when it was supplied by C

= 2% of the total notebooks

= 2/100

= 0.02

Probability that the defective notebook is supplied by A is denoted by P(E1/A).

Bayes' Theorem is used to find P(E1/A) which is given by the formula:

P(E1/A) = P(E1) × P(A/E1) / P(E1) × P(A/E1) + P(E2) × P(A/E2) + P(E3) × P(A/E3)

=> P(E1/A) = 0.25 × 0.05 / 0.25 × 0.05 + 0.35 × 0.04 + 0.4 × 0.02

=> P(E1/A) = 0.0125 / 0.0125 + 0.014 + 0.008

=>P(E1/A) = 0.0125 / 0.0345

=> P(E1/A) = 25/69

Therefore, the probability that the defective notebook was supplied by A is 25/69.