Conditional Probability: Distinction between $P(Fri|Defective)$ and $P(Defective|Fri)$ 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?
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.d = sum(.2*c(.08, .04, .04, .04, .12));  p.d
## 0.064


$P(F|D) = P(D \cap F)/P(D) = .024/.064 = 0.375.$ 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.
This problem's just a simple application of Bayes's theorems. We want to know $ P(Friday | defective)$ and we already have the inverse conditional probabilities e.g. $P(defective| Monday)$, $P(defective| Tuesday)$, etc.  The prior probability is 0.2 (this is written as:  &quot;20% of production on each weekday&quot;, 0.2 * 5 = 1): there are five days on the work week and we are indifferent to which day we choose.
Bayes Theorem:
$P(A|B) = \frac{P(B|A)*P(A)} {P(B)}$
$P(Defective| Friday) = P(B|A) = .12 $,  makes the first term multiplied by :
$P(A) = P(Friday) = P(any other day) = 0.2$, which is the prior probability.
In the denominator, $P(B)$ is the sum of the joint probabilities each one multiplied by the prior probability :
$ P(defective | Monday)P(Monday) + P(defective|tuesday)P(Tuesday)...+ etc., (0.08 + 0.04 + 0.04 ...) * 0.2.$
Since the prior probabilities are equal:
$$ P(Friday | defective) = \frac {0.12 * 0.2}{0.32 * 0.2} = 0.375 $$