Conditional probability formula |Bayes Theorem|Total Probability Law

Introduction

• Probability is the measure of the likeliness that an event will occur. Probability is quantified as a number between 0 and 1 (where 0 indicates impossibility and 1 indicates certainty)
  
• We have read about Terms related to Probability,Algebra of events operations< ,Classical Probability
in Previous class
• We have also read also Addition Theorems on Probability in previous classes
• Now we will learn about Conditional Probability

What is Conditional Probability

Let E and F are two events of the random experiments. Probability of event A happening give the condition event F has happened is called Conditional probability
So conditional probability of E given F has happened is P(E | F).
Similarly, we can define P (F| E).
We can think these conditional probabilities in these ways also
P (E|F) -> Probability of occurrence of E given F has happened
Or
P(E|F) -> Probability of occurrence of E when event F is taken as sample space

Multiplication Theorem on Probability|Conditional probability formula

(1) Let E and F are two events of a random experiments
Then (P (E ∩ F) = P(E) P(F|E) if P(E) ≠0
Or
 P (E ∩  F) = P(F) P(E|F) if P(F) ≠0
Proof:
Let S be the sample space and it contains n elementary events
Let a, b, c is the number of elementary event in E, F and (E ∩  F)
Then
P(E) = a/n
P(F) = b/n
P(E ∩  F)= c/n
Now
P(E |F)  =c/b
P(F |E) = c/a
Now
P(E ∩  F)= c/n
=(c/b)  (b/n)
= P(F) P( E|F) Or
P(E ∩  F)= c/n
=(c/a)  (a/n)
= P(E) P( F|E) Hence Proved
Now
P (E ∩ F) = P(E) P(F|E) if P(E) ≠0
Or
 P (E ∩ F) = P(F) P(E|F) if P(F) ≠0
Can be rewritten as
P(F|E) = P (E ∩  F) / P(E)
Or
P(E|F) = P (E ∩ F) / P(F)
2) So far we have proved all about dependent events
What are Independent Events
Events can be "Independent", meaning each event is not affected by any other events. We can say that if the occurrence or non-occurrence of one does not affect the probability of the occurrence or non-occurrence of the other
 Two events E and F are independent then
P(E|F) = P(E) and P(F|E) = P(F).
So P (E ∩ F) = P(E)P(F).
 We can extend this n number of events
Also for independents events
P (E ∪ F) = P(E) + P(F) – P (E  ∩ F)
P (E ∪ F) = P(E) + P(F) – P(E) P(F)
                  = 1 – P(E^c) P(F^c)

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