Bayes' Theorem
Bayes' theorem (or Bayes' Law and sometimes Bayes' Rule) is a direct application of conditional probabilities. The probability P(AB) of "A assuming B" is given by the formula
P(AB) = P(A∩B) / P(B) 
which for our purpose is better written as
P(A∩B) = P(AB)·P(B). 
The left hand side P(A∩B) depends on A and B in a symmetric manner and would be the same if we started with P(BA) instead:
P(BA)·P(A) = P(A∩B) = P(AB)·P(B). 
This is actually what Bayes' theorem is about:
(1)  P(BA) = P(AB)·P(B) / P(A). 
Most often, however, the theorem appears in a somewhat different form
(1')  P(BA) = P(AB)·P(B) / (P(AB)P(B) + P(AB)P(B)), 
where B is an event complementary to B:
This is because

and, since A∩B and A∩B are mutually exclusive,

More generally, for a finite number of mutually exclusive and exhaustive events H_{i}
H_{k} ∩ H_{m} = Φ, for k ≠ m and H_{1} ∪ H_{2} ∪ ... ∪ H_{n} = Ω, 
Bayes' theorem states that
P(H_{k}A) = P(AH_{k}) P(H_{k}) / ∑_{i} P(AH_{i}) P(H_{i}),
where the sum is taken over all i = 1, ..., n.
We shall consider several examples.
Example 1. Monty Hall Problem. [Havil, pp. 6163]
Let A, B, C denote the events "the car is behind door A (or #1)", "the car is behind the door B (or #2)", "the car is behine the door C (or #3)." Let also M_{A} denote the event of Monty opening door A, etc.
You are called on stage and point to door A, say. Then

Since A, B, C are mutually exclusive and exhaustive,

Now you are given a chance to switch to another door, B or C (depending on which one remains closed.) If you stick with your original selection (A),

However, if you switch,

You'd be remiss not to switch.
Example 2. Sick Child and Doctor. [Falk, p. 48]
A doctor is called to see a sick child. The doctor has prior information that 90% of sick children in that neighborhood have the flu, while the other 10% are sick with measles. Let F stand for an event of a child being sick with flu and M stand for an event of a child being sick with measles. Assume for simplicity that
A wellknown symptom of measles is a rash (the event of having which we denote R).
Upon examining the child, the doctor finds a rash. What is the probability that the child has measles?
Example 3. Incidence of Breast Cancer. [von Savant, pp. 103104]
In a study, physicians were asked what the odds of breast cancer would be in a woman who was initially thought to have a 1% risk of cancer but who ended up with a positive mammogram result (a mammogram accurately classifies about 80% of cancerous tumors and 90% of benign tumors.) 95 out of a hundred physicians estimated the probability of cancer to be about 75%. Do you agree?
References
 R. B. Ash, Basic Probability Theory, Dover, 2008
 R. Falk, Understanding Probability and Statistics, A K Peters, 1993
 J. Havil, Impossible?, Princeton University Press, 2008
 M. vos Savant, The Power of Logical Thinking, St. Martin's Press, NY 1996
 What Is Probability?
 Intuitive Probability
 Probability Problems
 Sample Spaces and Random Variables
 Probabilities
 Conditional Probability
 Dependent and Independent Events
 Algebra of Random Variables
 Expectation
 Probability Generating Functions
 Probability of Two Integers Being Coprime
 Random Walks
 Probabilistic Method
 Probability Paradoxes
 Symmetry Principle in Probability
 Nontransitive Dice
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Copyright © 19962017 Alexander Bogomolny
Sick Child and Doctor (Solution)
P(MR)  = P(RM) P(M) / (P(RM) P(M) + P(RF) P(F))  
= .95 × .10 / (.95 × .10 + .08 × .90)  
≈ 0.57. 
Which is nowhere close to 95% of P(RM).
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Incidence of Breast Cancer (Solution)
Introduce the events:
P   mammogram result is positive,  
B   tumor is benign,  
M   tumor is malignant. 
Bayes' formula in this case is
P(MP)  = P(PM) P(M) / (P(PM) P(M) + P(PB) P(B))  
= .80 × .01 / (.80 × .01 + .10 × .99)  
≈ 0.075  
≈ 7.5%. 
A far cry from a common estimate of 75%.
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