One box contains 5 products from machine A and 3 from machine B. It is known from past experience that 1% and 2% of the products made by each machine, respectively, are defective. Suppose that two products are randomly selected from the box. (a) What is the probability that one of them is defective? (b) If two products were chosen randomly and found to be defective, what is the probability that they were made by machine B?
Answers
Step-by-step explanation:
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
Let, E1 andE2
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 (E1)=60%=53
Probability of items produced by machine B,P (2)=40%=52
And,
Probability that machine A produced defective items, P(E1x)=2%=2/100
Probability that machine B produced defective items, P(E2x)=1%=1/100
So, the probability that randomly selected items was from machine A is given by P(xE1)
Now, Apply Bayes' Theorem
P(xE1)=
P(E1)P(E1x)+P(E2)P(E2x)P(E1)P(E1x)=52×1001+53×100253×1002=2+56=116
Hence, the required probability of machine A is
116
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