Bayes theroem with proof
Answers
Answer:
In probability theory and statistics, Bayes' theorem (alternatively Bayes's theorem, Bayes's law or Bayes's rule) describes the probability of an event, based on prior knowledge of conditions that might be related to the event.
Step-by-step explanation:
Example of Bayes' Theorem
P(A) – the probability that the stock price increases by 5%
P(B) – the probability that the CEO is replaced.
P(A|B) – the probability of the stock price increases by 5% given that the CEO has been replaced.
P(B|A) – the probability of the CEO replacement given the stock price has increased by 5%.
MARK ME AS BRANILIST
Step-by-step explanation:
Let AA and BB be two events and let P(A|B)P(A|B) be the conditional probability of AA given that BB has occurred. Then Bayes' theorem states that:
P(B|A)=P(A|B)P(B)P(A)=P(A|B)P(B)P(A|B)P(B)+P(A|Bc)P(Bc)(1)
(1)P(B|A)=P(A|B)P(B)P(A)=P(A|B)P(B)P(A|B)P(B)+P(A|Bc)P(Bc)
In other words, Bayes' theorem gives us the conditional probability of BB given that AA has occurred as long as we know P(A|B)P(A|B), P(A|Bc)P(A|Bc) and P(B)=1−P(Bc)P(B)=1−P(Bc).
The proof of this equation is quite simple. First, consider the following facts:
P(A|B):=P(A∩B)P(B)⟹P(A∩B)=P(A|B)P(B)(2)
(2)P(A|B):=P(A∩B)P(B)⟹P(A∩B)=P(A|B)P(B)
P(A)===P((A∩B)∪(A∩Bc))P(A∩B)+P(A∩Bc)P(A|B)P(B)+P(A|Bc)P(Bc)(3)
P(A)=P((A∩B)∪(A∩Bc))=P(A∩B)+P(A∩Bc)(3)=P(A|B)P(B)+P(A|Bc)P(Bc)
where on equation (3)(3) the fact that A∩BA∩B and A∩BcA∩Bc are mutually exclusive events was used. Therefore, since A∩B=B∩AA∩B=B∩A:
P(B|A)===P(B∩A)P(A)P(A|B)P(B)P(A)P(A|B)P(B)P(A|B)P(B)+P(A|Bc)P(Bc)(4)
P(B|A)=P(B∩A)P(A)=P(A|B)P(B)P(A)(4)=P(A|B)P(B)P(A|B)P(B)+P(A|Bc)P(Bc)
as we wanted to prove. We can also prove a very interesting formula using Bayes' theorem. From equation (4)(4) (used with both BB and BcBc) and equation (3)(3), we have:
P(B|A)+P(Bc|A)=P(A|B)P(B)P(A)+P(A|Bc)P(Bc)P(A)=P(A)P(A)=1(5)
(5)P(B|A)+P(Bc|A)=P(A|B)P(B)P(A)+P(A|Bc)P(Bc)P(A)=P(A)P(A)=1
so we obtain:
P(B|A)=1−P(Bc|A)(6)
(6)P(B|A)=1−P(Bc|A)