Question

In a bolt factory, machines A, B, and C manufacture 25%, 35% and 40% of the total output, respectively. Of their total of the output, 5%, 4% and 2% respectively are defective bolts. A bolt is drawn at random and is found to be defective. What is the probability that it is manufactured by machine B?

Answer 1

4/35

Answer 2

\[P(A\ defective)=0.25\times0.05\]
\[P(B\ defective)=0.35\times0.04\]
\[P(C\ defective)=0.4\times0.12\]
The probability that the defective bolt is made by machine B is given by:
\[\large P(made\ by\ B)=\frac{P(B\ defective)}{P(A\ defective)+P(B\ defective)+P(C\ defective)}\]

Answer 3

140/100

Answer 4

0.14/0.2005

Answer 5

Not really. Lets look at the numerator value:
\[\large P(B\ defective)=0.35\times0.04=what\ is\ your\ result?\]

Answer 6

0.014/0.0745

Answer 7

0.014

Answer 8

The numerator is correct now. What are you results for:
P(A defective) = ?
P(C defective) = ?

Answer 9

0.0125,0.0008

Answer 10

P(C defective)=0.4×0.12 u wrote there instead of 0.02.thats why i got confused

Answer 11

0.014/0.0345

Answer 12

P(A defective) = 0.0125
P(B defective) = 0.0140
P(C defective) = 0.0080
Sorry for typo earlier.

0.014/0.0345 is the correct fraction. Now you can finally calculate the result.

Answer 13

so 28/69

Answer 14

why we are multiplying the output with defective.i thought defective is given directly

Answer 15

0.35*0.04 why?.do u understand my question

Answer 16

4/35 itself show it is manufactured by b

Answer 17

Let event B be that a bolt came from machine B. Let event D be the bolt is defective.
P(B) = 0.35
P(D) = 0.04
The probability of B and D is:
\[P(B \cap D)=P(B) \times P(D)=0.35\times0.04\]

Answer 18

nice:-)

Answer 19

You're welcome :)

Answer 20

when we use this formula.

Answer 21

The formula applies when the two events are independent.