Question:
The factory quality control department discovers that the conditional probability of making a manufacturing mistake in its precision ball bearing production is 4% on Tuesday, 4% on Wednesday, 4% on Thursday, 8% on Monday, and 12% on Friday.
The Company manufactures an equal amount of ball bearings (20%) on each weekday.
What is the probability that a defective ball bearing was manufactured on a Friday?
My solution:
Given, P(defective|Monday)=0.08
P(defective|Tuesday)=0.04
P(defective|Wednesday)=0.04
P(defective|Thursday)=0.04
P(defective|Friday)=0.12
Now, the question asks for P(Friday|Defective)? Is my logic correct? Could someone help in how to finish the problem?
Best Answer
So far you have properly identified $P(D|M), \dots, P(D|F).$ But you need to look at Bayes' Theorem and figure out how to get the 'inverse' or 'posterior' probability $P(F|D).$ Here is an outline; you should match the steps with Bayes' Theorem.
Notice that $P(D \cap F),\, P(D|F)$ and $P(F|D)$ are all probabilities involving events $D$ and $F.$ But they refer to different populations: $P(D \cap F)$ refers to the entire weekly production of ball bearings; $P(D|F)$ refers to the population of ball bearings made on Friday; and $P(F|D)$ refers to the population of all defective ball bearings, asking what proportion of them were made on Friday.
$P(D \cap M) = P(M)P(D|M) = .2(.08) = .16.$ This uses what is sometimes called the 'general multiplication rule'.
$P(D) = P(D \cap M) + P(D \cap Tu) + \cdots + P(D \cap F) = 0.064.$ This uses what is sometimes called the 'law of total probability'.
Using R statisical software as a calculator:
$P(F|D) = P(D \cap F)/P(D) = .024/.064 = 0.375.$ This uses Bayes' Theorem.
While only 20% of all ball bearings are made on Friday, 37.5% of defective ball bearings were made on Friday.