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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
-------------
From before we had:
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(A_{1}|B) = P(B|A_{1}).P(A_{1})/Σ_{k}P(B_{k})
= P(B|A_{1}).P(A_{1})/Σ_{k}P(B|A_{k}).P(A_{k})
≡ P(A_{1} ∩ B)/Σ_{k}P(A_{k} ∩ B)
This is BAYES' THEOREM.
P(B) = Σ_{k}P(B_{k}) ≡ Σ_{k}P(B|A_{k}).P(A_{k}) ≡ Σ_{k}P(A_{k} ∩ 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(B_{1} ∩ B_{2}) = P(B_{2}|B_{1}).P(B_{1})
1 blue given 1 blue ( P(B_{1} ∩ B_{2}) ): (2/5)(1/4) = 1/10
1 red given 1 blue ( P(R_{1} ∩ B_{2}) ): (2/5)(3/4) = 3/10
1 blue given 1 red ( P(B_{1} ∩ R_{2}) ): (3/5)(2/4)^{*} = 3/10
1 red given 1 red ( P(R_{1} ∩ R_{2}) ): (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(B_{1} ∩ B_{2}) ): (2/5)(2/5) = 4/25
1 red given 1 blue ( P(R_{1} ∩ B_{2}) ): (2/5)(3/5) = 6/25
1 blue given 1 red ( P(B_{1} ∩ R_{2}) ): (3/5)(2/5) = 6/25
1 red given 1 red ( P(R_{1} ∩ R_{2}) ): (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(B_{1}|B_{2}) = P(B_{2}|B_{1}).P(B_{1})/P(B_{2})
Where,
P(B_{1}) is the PRIOR, the initial probability of B_{1}.
P(B_{1}|B_{2}) is the POSTERIOR, the probability of B_{1} after B_{2}.
P(B_{2}|B_{1})/P(B_{2}) is the support B provides for B_{1}.
But P(B_{2}) = 1/10 + 3/10 from answer 1. Therefore,
P(B_{1}|B_{2}) = (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!.