A factory has three machines, A, B, and C, for

l2yanhatal0n

l2yanhatal0n

Answered question

2022-02-11

A factory has three machines, A, B, and C, for producing items. Machine A produces 50%, B produces 30%, and C produces 20%. If a randomly selected item from the factory's output is found to be defective, what is the probability that B made it?

Answer & Explanation

Ijezid8t

Ijezid8t

Beginner2022-02-12Added 13 answers

Step 1
Prior information
Probability of an item produced by Machine A P(A)=50%=0.5
Probability of an item produced by Machine B P(B)=30%=0.3
Probability of an item produced by Machine C P(C)=20%=0.2
Considering additional information
The likelihood that the product made by A is flawed is:
P(DA)=3%=0.03
The likelihood that the product made by B is flawed is:
P(DB)=2%=0.02
The likelihood that the product made by C is flawed is:
P(DC)=1%=0.01
Given the item is defective, probability of it being produced by B
P(BD)=P(B)×P(DB)[P(B)×P(DB)]+[P(A)×P(DA)]+[P(C)×P(DC)]
=0.3×0.02[0.3×0.02]+[0.5×0.03]+[0.2×0.01]
0.0060.006+0.015+0.002=0.0060.023=0.26
Given the item is defective, probability of it being produced by B=0.26

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