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Conditional Probability
-----------------------

If A and B are DEPENDENT events.  The probability of B given A,
P(B|A), is given by:

P(B|A) = P(A ∩ B)/P(A) ... 1.

Note:  If the events are INDEPENDENT then:

P(A ∩ B) = P(A).P(B)

Rearranging 1. we get:

P(A ∩ B) = P(B|A).P(A)

Bayes Theorem
-------------

P(A ∩ B) = P(B|A).P(A)

We can also write:

P(B ∩ A) = P(A|B).P(B)

Since P(A ∩ B) = P(B ∩ A),

P(B|A).P(A) = P(A|B).P(B)

Or,

P(A|B) = P(B|A).P(A)/P(B)

Where,

P(A) is the PRIOR, the initial probability of A.
P(A|B) is the POSTERIOR, the probability of A after B.
P(B|A)/P(B) is the support B provides for A.

From the above Tree diagram we can see that:

P(B) = P(A ∩ B) + P(~A ∩ B)

= P(A).P(B|A) + P(~A).P(B|~A)

Therefore,

P(A|B) = P(B|A).P(A)/{P(A).P(B|A) + P(~A).P(B|~A)}

We can generalize this to:

P(A1|B) = P(B|A1).P(A1)/ΣkP(Bk)

= P(B|A1).P(A1)/ΣkP(B|Ak).P(Ak)

≡ P(A1 ∩ B)/ΣkP(Ak ∩ B)

This is BAYES' THEOREM.

P(B) = ΣkP(Bk) ≡ ΣkP(B|Ak).P(Ak) ≡ ΣkP(Ak ∩ B) is the LAW OF TOTAL
PROBABILITY.

Bayes' Theorem calculates the probability that an event (A) occurs
given knowledge of another event (B).  It can also be understood
as a way of understanding how the probability that a theory is true
is affected by a new piece of evidence.

It is used for many purposes, including detecting faults, surveillance,
military defence, search-and-rescue operations, medical screening
and even email spam filters.

Example 1:

Consider 5 marbles in a bag:  2 blue and 3 red.  What is the
probability of:

1.  Pulling 2 blue marbles without replacement?
2.  At least one blue marble being pulled?
3.  Pulling 2 blue marbles with replacement?
4.  The first marble being blue given that the second marble is
blue with replacement.

1.  These are DEPENDENT events.  To calculate the probability of
pulling a blue marble given that the first marble picked was blue,
we use the formula:

P(B1 ∩ B2) = P(B2|B1).P(B1)

1 blue given 1 blue ( P(B1 ∩ B2) ):  (2/5)(1/4) = 1/10

1 red given 1 blue ( P(R1 ∩ B2) ):   (2/5)(3/4) = 3/10

1 blue given 1 red ( P(B1 ∩ R2) ):   (3/5)(2/4)* = 3/10

1 red given 1 red ( P(R1 ∩ R2) ):    (3/5)(2/4) = 3/10

* Remember a blue marble wasn't picked so there must still be 2.

The probabability of 2 blue is 1/10.

2.  The probability of at least 1 blue is 1/10 + 3/10 + 3/10 = 7/10.

3.  These events are INDEPENDENT and we get:

1 blue given 1 blue ( P(B1 ∩ B2) ):  (2/5)(2/5) = 4/25

1 red given 1 blue ( P(R1 ∩ B2) ):   (2/5)(3/5) = 6/25

1 blue given 1 red ( P(B1 ∩ R2) ):   (3/5)(2/5) = 6/25

1 red given 1 red ( P(R1 ∩ R2) ):    (3/5)(3/5) = 9/25

The probabability of 2 blue is 4/25.  The probability of at least
1 blue is 4/25 + 6/25 + 6/25 = 14/25.

4.  To calculate the probability that the first marble was blue
given that the second marble picked was blue, we use Bayes'
formula:

P(B1|B2) = P(B2|B1).P(B1)/P(B2)

Where,

P(B1) is the PRIOR, the initial probability of B1.
P(B1|B2) is the POSTERIOR, the probability of B1 after B2.
P(B2|B1)/P(B2) is the support B provides for B1.

But P(B2) = 1/10 + 3/10 from answer 1.  Therefore,

P(B1|B2) = (2/5)(1/4)/{1/10 + 3/10}   ... dotted red lines

= 1/4

So the probability that a blue was picked initially given that a
another blue was picked the second time is 'modified' from
the prior value of 2/5 to 1/4!

Example 2:

1% of women have breast cancer.  A woman with breast cancer
has a 90% chance of testing positive while a woman without
has a 5% chance (false positive).   What is the probability a
women has breast cancer given that she had a positive test?

P(C|+) = P(+|C).P(C)/P(+)

+ given cancer ( P(B ∩ B) ):  (0.01)(0.90) = 0.0090

- given cancer ( P(R ∩ B) ):   (0.01)(0.10) = 0.0010

+ given no cancer  ( P(B ∩ R) ):   (0.99)(0.10) = 0.0990

- given no cancer ( P(R ∩ R) ):    (0.99)(0.9)  = 0.8910

P(C|+) = 0.0090/{0.0090 + 0.0990}

= 0.0090/0.1080

= 0.08333

So the probability of having cancer given that the test is
positive is actually quite low!.