Question

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?

Answer 1

the answer of the question is 0.1

Answer 2

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

.