A factory has two Machines I and II. Machine I and II produce 30% and 70% of items
respectively. Further, 3% of items produced by Machine I are defective and 4% of
items random. If the drawn item is defective, find the probability that it was produced
by Machine II.
Answers
Answer:
Let E
1
and E
2
be the respective events of items produced by machines A and B. Let X be the event that the produced item was found to be defective.
∴ Probability of items produced by machine A, P(E
1
)=60% =
5
3
Probability of items produced by machine B, P(E
2
)=40% =
5
2
Probability that machine A produced defective items, P(X∣E
1
)=2% =
100
2
Probability that machine B produced defective items, P(X∣E
2
)=1% =
100
1
The probability that the randomly selected item was from machine B, given that it is defective, is given by P(E
2
∣X).
By using Baye's theorem, we obtain
P(E
2
∣X)=
P(E
1
)⋅P(X∣E
1
)+P(E
2
)⋅P(X∣E
2
)
P(E
2
)⋅P(X∣E
2
)
=
5
3
⋅
100
2
+
5
2
⋅
100
1
5
2
⋅
100
1
=
500
6
+
500
2
500
2
=
8
2
=
4
1
=0.25
Answer:
Consider the problem
Let, E
1
andE
2
be the respective events of items produced by machine A and B.
And let x be the event that the produced item was found to be defective.
Therefore,
Probability of items produced by machine A,P (E
1
)
=60%=
5
3
Probability of items produced by machine B,P (E
2
)
=40%=
5
2
And,
Probability that machine A produced defective items, P(
E
1
x
)
=2%=
100
2
Probability that machine B produced defective items, P(
E
2
x
)
=1%=
100
1
So, the probability that randomly selected items was from machine A is given by P(
x
E
1
)
Now, Apply Bayes' Theorem
P(
x
E
1
)=
P(E
1
)P(
E
1
x
)+P(E
2
)P(
E
2
x
)
P(E
1
)P(
E
1
x
)
=
5
2
×
100
1
+
5
3
×
100
2
5
3
×
100
2
=
2+5
6
=
11
6
Hence, the required probability of machine A is
11
6