Math, asked by jj2002, 2 months ago

A manufacturing firm produces pipes in two plants I and II with daily production 1500 and 2000 pipes respectively. The

fraction of defective pipes produced by two plants I and II are 0.006 and 0.008 respectively. If a pipe selected at random

from that day’s production is found to be defective, what is the chance that it has come from plant I or plant II?​

Answers

Answered by mohammedayubjafar111
0

Answer:

Let us consider the probabilities of production

P(A)=

3500

500

=

7

1

P(B)=

3500

1000

=

7

2

P(C)=

3500

2000

=

7

4

And,

Probability of defective pipes

By plant A=P(A/E)=0.005

By plant B=P(B/E)=0.008

By plant C=P(C/E)=0.010

Now,

from Bayes theorem

Probability of defective pipe from first plant

P(E/A)=

[P(A).P(A/E)+P(B).P(B/E)+P(C).P(C/E)]

[P(A)×P(A/E)]

=

[

7

1

×0.005+

7

2

×0.008+

7

3

×0.010]

[

7

1

×0.005]

=

[

7

0.005

+

7

2(0.008)

+

7

3(0.010)

]

[

7

0.005

]

=

0.005+0.016+0.040

0.005

=

0.061

0.005

=

61

5

So, the probability is from First factory is

61

5

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