Question

In bolt factory , machine A ,B ,and C manufacturer 25, 35 and 40

Percent of the total output , 5, 4 and 2 percent , respectively are

defective bolts.A bolts is selected at random and found to be defective.

What is the probability that the bolt came from machine A , B and C ?

Answer 1

here, $P(A)= \frac {25}{100} ​$, $P(B)= \frac {35}{100} ​$, $P(C)= \frac {40}{100} ​$

$P(D/A)= \frac {5}{100} ​$, $P(D/B)= \frac {4}{100} ​$, $P(D/C)= \frac {2}{100} ​$

where D denotes defective bolts

$P(D)=P(A).P( D/A )+P(B).P(D/B )+P(C).P( D/C)$

$P(D)=\frac {25}{100}.\frac {5}{100}+\frac {35}{100} ​ \frac {4}{100} +\frac {40}{100} ​ \frac {2}{100}$ $= 0.0345$

$P(A/D)=\frac {P(A).P( D/A)} {P(A).P( D/A)+P(B).P(D/B )+P(C).P( D/C)}$

$P(A/D)=\frac {\frac{25}{100}. \frac {5}{100} } {0.0345}=\frac {25}{69}$

$P(B/D)=\frac {P(B).P( D/B)} {P(A).P( D/A)+P(B).P(D/B )+P(C).P( D/C)}$

$P(B/D)=\frac {\frac{35}{100}. \frac {4}{100} } {0.0345}=\frac {28}{69}$

$P(C/D)=\frac {P(C).P( D/C )} {P(A).P( D/A )+P(B).P(D/B)+P(C).P( D/C)}$

$P(C/D)=\frac {\frac{40}{100}. \frac {2 }{100} } {0.0345}=\frac {16}{69}$