Essential Maths for ML – Part 3
So let’s consider A1, A2, A3, and A4 be the mutually
exclusive and exhaustive event of a random experiment. Let B be the common
event i.e. the event B is made-up of 4- mutually exclusive and exhaustive
events.
P(B) = P(A1⋂B)
+ P(A2⋂B) + P(A3⋂B) + P(A4⋂B)
P(B) = Σ P(Ai⋂B)…………(1)
We already know from the concept of the conditional probability
that
P(A1⋂B) = P(B) * P(A1/B)
P(A1/B) = P(A1⋂B) / P(B)………(2)
Replacing the value of P(B) from the eq2 we can say that
P(A1/B) = P(A1⋂B) / Σ P(Ai⋂B)
P(A1/B) = P(A1) * P(B/A1) / Σ P(Ai⋂B)
So the Bayes Theorem states that if A1, A2, A3……….An, are n
mutually exclusive and exhaustive events with prior probabilities P(A1), P(A2),
P(A3)……………P(An) and B be the common event for which the conditional probabilities
of the probability of occurrence of B given A1, B given A2 ………….B given An will
be P(B/A1), P(B/A2), P(B/A3)………..P(B/An) respectively then the posterior
probability of occurrence of A1 given that B has already occurred is given by :
P(A1/B) = P(A1/B) * P(B) / Σ P(Ai) * P(B/Ai)……..i=1 to n
Let’s understand this theorem with an example.
The probabilities that Mr. A, Mr. B and Mr. C will get the promotion
are 0.40, 0.35 and 0.25 respectively. The
probabilities that they will introduce new business are 0.10, 0.15, and 0.20
respectively. What is the probability that Mr B introduces a new business by
getting the promotion?
· Let A1 be
the event for Mr. A getting the promotion
· Let A2 be
the event for Mr. B getting the promotion
· Let A3 be
the event for Mr. C getting the promotion
· Let B be
the event that a new business was introduced
So according to the above definitions we can say that
P(A1) = 0.40, P(A2) = 0.35, P(A3) = 0.25,
P(B/A1) = 0.10, P(B/A2) = 0.15 and P(B/A3) = 0.20
We need to find out the value of P(A2/B)
Event Ai |
Prior Probability P(Ai) |
Conditional
Probability P(B/Ai) |
Joint Probability
P(Ai⋂B) = P(Ai)* P(B/Ai) |
Posterior
probability P(Ai/B) = P(Ai⋂B) / Σ
P(Ai) * P(B/Ai) |
A1 |
0.40 |
0.10 |
0.0400 |
0.0400/0.1425
= 0.2807 |
A2 |
0.35 |
0.15 |
0.0525 |
0.0525/0.1425
= 0.3684 |
A3 |
0.25 |
0.20 |
0.0500 |
0.0500/0.1425
=0.3509 |
Sum |
1.0 |
0.45 |
0.1425 |
|
I hope you will find this blog useful and informative. In my next blog we will continue discussing the mathematical concepts required to understanding the machine learning models. Please do not forget to follow me on my blog page.
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