Random Signals for Engineers using MATLAB and Mathcad
Copyright  1999 Springer Verlag NY
Example 1.7
Let us assume that three machines are maufacturing parts . For every bin we fill with 100 parts
machine A contributes 20 parts
machine B contributes 30 parts
machine C contributes50 parts
In addition, we know that for every 100 parts manufactured that each machine manufacters
machine A makes 6 defective parts
machine B makes 7 defective parts
machine C makes 8 defective parts
SOLUTION: Let us first compute the probability that a part selected from the bin is defective. We can
define the following probabilities:
Pa=0.2;Pb=0.3;Pc=0.5;
[Pa Pb Pc]
ans =
0.2000
0.3000
0.5000
Pda=0.06;Pdb=0.07;Pdc=0.08;
[Pda Pdb Pdc]
ans =
0.0600
0.0700
0.0800
We have defined PdA to be the P[defective part from bin | machine A made it] = P dA. Using the law of
total probability we have
Pd=Pda*Pa+ Pdb*Pb + Pdc*Pc
Pd =
0.0730
Now we may determine given that we have selected a part and it is defective what is the probability that is
was manufactured by machine A, B or C. This may be computed using the Equation 1.7-6. This is said to
be the a posterori probability because we are computing the probability of an event after we have
knowledge of a prior event occuring. Defining P[machine A made it | a defective part was taken from the
bin] = PAd we have
Pad=Pda*Pa/Pd;
[Pad Pbd Pcd]
ans =
0.1644
Pbd=Pdb*Pb/Pd;
0.2877
Pcd=Pdc*Pc/Pd;
0.5479
We, of course, may verify that the sum of all the conditional probabilities is unity, because one machine
had to make the defective part.
Pad + Pbd +Pcd
ans =
1