1. ## independent (probability theory)

a factory produce a product that can have 3 different faults
Fault: A,B,C
The fault is independent (probability theory) of each other.

X: number of faults that one product can have.

Determine the density function PX(x) if P(A)= 0,20, P(B)=0,05, P(B)=0,10 ?

2. Originally Posted by sf1903
a factory produce a product that can have 3 different faults
Fault: A,B,C
The fault is independent (probability theory) of each other.

X: number of faults that one product can have.

Determine the density function PX(x) if P(A)= 0,20, P(B)=0,05, P(B)=0,10 ?
$P(X=0)=(1-0.2)(1-0.05)(1-0.10)$

that is there is no fault A, no fault B and no fault C

$p(X=1)=0.2(1-0.05)(1-0.10)+(1-0.2)0.05(1-0.10)+(1-0.2)(1-0.05)0.10$

that is it is the sum of the probabilities that there is a fault A but no B or C, that there is fault B but no A or C and there is fault C but no A or B

$p(X=2)= ...$

CB

3. Originally Posted by CaptainBlack
$P(X=0)=(1-0.2)(1-0.05)(1-0.10)$

that is there is no fault A, no fault B and no fault C

$p(X=1)=0.2(1-0.05)(1-0.10)+(1-0.2)0.05(1-0.10)+(1-0.2)(1-0.05)0.10$

that is it is the sum of the probabilities that there is a fault A but no B or C, that there is fault B but no A or C and there is fault C but no A or B

$p(X=2)= ...$

CB

P(X=3)=0,2*0,05*0,10= 0,001
We get P(X=2)=0,032 and P(X=3)=0,001 is it right?