1. A factory has 3 machines X, Y and Z producing 1000, 2000 and 3000 bolts per day respectively. The machine X produces 1percentdefective bolts, Y produces 1.5percent and Z produces 2percentdefective bolts. At the end of the day, a bolt is drawn at random and is found to be defective. What is the probability that this defective bolt has been produced by machine X?
Answers
Answer:
Step-by-step explanation:
P(X) = P ( that the machine X produce
bolts)
= 1000/6000
= 1/6
P(Y) = P ( that the machine Y produce
bolts)
= 2000/6000
= 1/3
P(Z) = P ( that the machine Z produce
bolts)
= 3000/6000
= 1/2
Let A be the event that the choosen bolts is defective.
P( A/X) = P( that defective bolts from
the machine X )
= 0.01
P(A/Y) = P ( that defective bolts from
the machine Y )
= 0.015
P(A/Z) = P ( that defective bolts from
the machine Z )
= 0.02
We have to find P(X/A)
Hence by Baye's theorem, we get
P(X/A) = P(X)P(A/X)
—————————————
P(X)P(A/X)+P(Y)P(A/Y)+
P(Z)P(A/Z).
(1/6)(0.01)
= —————————————————
(1/6)(0.01)+(1/3)(0.015)+(1/2) (0.02)
0.01
= ——————————
0.01+0.03+0.06
= 0.01/0.1
= 1/10
.