Q.No. 01: 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
Answer:
Step-by-step explanation:
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 is the answer
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Answer:
plants A, B and C with daily production of 500, 1000 and 2000 units respectively. The fractions of defective steel pipes output produced by the plant A, B and C are respectively 0.005, 0.008 and 0.010. If a pipe is selected from a day’s total production and found to be defective, find out the probability that it came from the first plant.
Solution :
We have probability of production,
P(A) = 500/3500 = 1/7,
P(B) = 1000/3500 = 2/7,
P(C) = 2000/3500 = 4/7.
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.
Therefore by Baye’s Theorem, probability of defective pipe from first plant
= P(E/A) = [P(A)×P(A/E)]/[P(A).P(A/E) + P(B).P(B/E) + P(C).P(C/E)]
= [1/7×0.005]/[1/7×0.005 + 2/7×0.008 + 4/7×0.010]
= 0.005/[0.005 + 0.016 + 0.040]
= 0.005/0.061 = 5/61[Ans.]