Probability question...?

Question

(Please answer fully...thanks) Here's the question neways:-

A company produces MP3 players at three different factories.

Factory A produces twice as many MP3 players as factory B.

Factory C produces as many MP3 players as factory A and factory B combined. The probablity that an MP3 player is faulty is 0.05 if it is produced by factory A. For factories B and C, the probablity that an MP3 player produced by factory is faulty is 0.04 and 0.03, respectively.

An Mp3 player is chosen at random from all those produced by the company

1)Find the probablity that the chosen one was produced by factory B.

2)What is the proabablity the the chosen one was faulty.

3)Given that the MP3 player is faulty, what is the probablity that it was produced at facotry A?

4)Given that the MP3 player is not faulty, what is the probablity that it was produced at factory C?

Answer 1

A:B:C = 2:1:3

p(A-ok) = 2/6 * 95/100 = 190/600

p(A-f) = 2/6 * 5/100 = 10/600

p(B-ok) = 1/6 * 96/100 = 96/600

p(B-f) = 1/6 * 4/100 = 4/600

p(C-ok) = 3/6 * 97/100 = 291/600

p(C-f) = 3/6 * 3/100 = 9/600

1) 1/6

2) 5/100

3) all faulty produkts from 600  

A=10, B=4 , C= 9

p=10/23

4) all "ok" produkts from 600

A=190, B=96 ,C=291

p =291/ 577

  

Answer 2

A=Factory A

B=Factory B

C= Factory C

F= faulty

Let us assume factory B produces x players. Then, factory A produces 2x players and factory C produces 2x+x=3x players

There are 6x players produced in all.

P(A)=2/6=2x/6x

P(B)=1/6=x/6x

P(C)=3/6=1/2=3x/6x

P(F/A)=0.05

P(F/B)=0.04

P(F/C)=0.03

1) P(B)=1/6

2) P(F) = P(A)P(F/A)+P(B)P(F/B)+P(C)P(F/C)

=(1/3)(0.05) + (1/6)(0.04) + (1/2)(0.03)

=0.01666+0.00666+0.015=0.03832

3)P(A/F) = P(A)P(F/A) /[P(A)P(F/A)+P(B)P(F/B)+P(C)P(F/C)]

=0.01666 / 0.03832 =0.4348

4)Find P(C/F) (given it 'is' faulty)

P(C/F)=P(C)P(F/C) / P(A)P(F/A) /[P(A)P(F/A)+P(B)P(F/B)+P(C)P(F/C)]

=0.015 / 0.03832=0.39144

So, the required probability is 1-0.39144 =0.60855