 # Conditional probability formula |Bayes Theorem|Total Probability Law

This Chapter we would be taking a look at the below topics
Conditional probability , Multiplication Theorem on 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

Example
Let take an example of a random experiment of Flip a coin twice.
S = {(H, H), (H, T), (T, H), (T, T)}.
Let F be the event of first flip to be H.
And let Me be the event for two flips are not both H
Then F = {(H, H), (H, T)}, P(F) = 1/2.
E = {(H, T), (T, T), (T, H)}, P(E) = 3/4.
Now If F event occurs, in order for event E to occur, the second flip has to be T.
P(E|F) = ½

## 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(Ec) P(Fc)

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