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# Bayes' Theorem
### Example
$1.$ We have two boxes.\
$~~~~~$a) The first contains two green balls and seven red balls\
$~~~~~$b) The second contains four green balls and three red balls.\
$2.$ Bob selects a ball by first choosing one of the two boxes at random.\
$3.$ He then selects one of the balls in this box at random.\
$4.$ If Bob has selected a red ball, what is the probability that he selected a ball from the first box?
Solution:
$1.$ Let\
$~~~~~$a) $E$ be the event that Bob has chosen a red ball\
$~~~~~$b) $\overline{E}$ be the event that Bob has chosen a green ball\
$~~~~~$c) $F$ be the event that Bob has chosen a ball from the first box\
$~~~~~$d) $\overline{F}$ be the event that Bob has chosen a ball from the second box.
$2.$ We want to find $p(F|E)$, the probability that the ball Bob selected came from the first box, given it is red
$~~~~~ p(F|E) = \dfrac{p(F \cap E)}{p(E)}$
## Bayes' Theorem
Suppose that $E$ and $F$ are events from a sample space $S$ such that $p(E) \neq 0$ and $p(F) \neq 0$. Then
$~~~~~ p(F|E) = \dfrac{p(E|F) \cdot p(F)}{p(E|F) \cdot p(F) + p(E|\overline{F}) \cdot p(\overline{F})}$
### Example
Suppose that one person in 100,000 has a particular rare disease for
which there is a fairly accurate diagnostic test. This test is correct 99.0%
of the time when given to a person selected at random who has the
disease; it is correct 99.5% of the time when given to a person selected
at random who does not have the disease. Given this information can
we find
$~~~~~$a) the probability that a person who tests positive for the disease has the disease?
$~~~~~$b) the probability that a person who tests negative for the disease does not have the disease?
Should a person who tests positive be very concerned that he or she
has the disease?
Solution:
Let
$~~~~~$a) $F$ be the event that a person selected at random has the disease
$~~~~~$b) $E$ be the event that a person selected at random tests positive for the disease.
We want to compute $p(F|E)$. To use Bayes' theorem to compute $p(F|E)$, we need to find $p(E|F)$, $p(E|\overline{F})$, $p(F)$, and $p(\overline{F})$.
We know that one person in 100,000 has this disease, so $p(F) = \dfrac{1}{100000} = 0.00001$ and $p(\overline{F}) = 1 - p(F) = 0.99999$.
Because a person who has the disease tests positive 99% of the time, we know that $p(E|F) = 0.99$.
This is the probability of a true positive, that a person with the disease tests positive. It follows that $p(\overline{E}|F) = 1 - p(E|F) = 1 - 0.99 = 0.01$; this is the probability of a false negative, that a person with the disease tests negative.
Furthermore, because a person who does not have the disease tests negative 99.5% of the time, we know that $p(\overline{E}|\overline{F}) = 0.995$. This is the probability of a true negative, that a person without the disease tests negative.
Finally, we see that $p(E|\overline{F}) = 1 - p(\overline{E}|\overline{F}) = 1 - 0.995 = 0.005$. This is the probability of a false positive, that a person without the disease tests positive.
The probability that a person who tests positive for the disease actually has the disease is $p(F|E)$. By Bayes' theorem, we have
$~~~~~ p(F|E) = \dfrac{p(E|F) \cdot p(F)}{p(E|F) \cdot p(F) + p(E|\overline{F}) \cdot p(\overline{F})}$
$~~~~~ p(F|E) = \dfrac{0.99 \cdot 0.00001}{0.99 \cdot 0.00001 + 0.005 \cdot 0.99999} \approx 0.002$
The probability that someone who tests negative for the disease does not have the disease is $p(F|E)$. By Bayes' theorem, we know that
$~~~~~ p(\overline{F}|\overline{E}) = \dfrac{p(\overline{E}|\overline{F}) \cdot p(\overline{F})}{p(\overline{E}|\overline{F}) \cdot p(\overline{F}) + p(\overline{E}|F) \cdot p(F)}$
$~~~~~ p(\overline{F}|\overline{E}) = \dfrac{0.995 \cdot 0.99999}{0.995 \cdot 0.99999 + 0.01 \cdot 0.00001} \approx 0.99998$
Consequently, 99.99999% of the people who test negative really do not have the disease.
In part (a) we showed that only 0.2% of people who test
positive for the disease actually have the disease. People
who test positive for the diseases should not be overly
concerned that they actually have the disease