Answer to Question #259274 in Statistics and Probability for MARYG

Question

An accounting company buys its computers from three different companies. Company X supplies 25% of the computers, company Y supplies 35% of the computers and company Z supplies the rest. From past experience you determine that 5% of company X’s computers produced are defective, 4% of company Y’s computers are defective and 3% of company Z’s computers are defective. One of the computers was reported as defective. By using Bayes' theorem or another method, what is the probability that the computer was supplied by Company X? 

Answer 1

Let us first define the events described in this question below.

Let X be the event that computers are manufactured by company X, Y be the event that computers are manufactured by company Y and Z be the event that computers are manufactured by company Z.

Also, Let D be the event that the computers manufactured are defective.

Define the probabilities,

$p(X)=25\%=0.25$

$p(Y)=35\%=0.35$

$p(Z)=40\%=0.40$

The next part of the question describes the conditional probabilities as shown below,

$p(D|X)=5\%=0.05$

$p(D|Y)=4\%=0.04$

$p(D|Z)=3\%=0.03$

We shall first determine the probability that a selected computer is defective using the law of total probability given as,

$p(D)=p(D|X)*p(X)+p(D|Y)*p(Y)+p(D|Z)*p(Z)$

$p(D)=(0.05*0.25)+(0.04*0.35)+(0.03*0.40)= 0.0385$

This question requires us to find, $p(X|D)$.

Using Bayes Theorem, this probability is given as, $p(X|D)=(p(D|X)*p(X))/(p(D|X)*p(X)+p(D|Y)*p(Y)+p(D|Z)*p(Z)), \\\,p(D)=p(D∣X)∗p(X)+p(D∣Y)∗p(Y)+p(D∣Z)∗p(Z)),$

This can also be written as

$p(X|D)=(p(D|X)*p(X))/p(D)$

$p(X|D)=(0.05*0.25)/0.0385=0.0125/0.0385= 0.3246753$

Therefore, given that the computer was defective, the probability that the computer was supplied by Company X is 0.3247(4 decimal places)