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Sets, Combinatorics, and Probability
At this point we have what might be called an addition rule and a multiplication rule for probability, loosely related to the addition principle and the multiplication principle for counting. If we want the probability of event E 1 or event
E 2 , that is, P(E 1 c E 2 ), we can add the respective probabilities—but only if the
events are disjoint. If we want the probability of event E 1 and event E 2 , that is,
P(E 1 d E 2 ), we can multiply the respective probabilities—but only if the events
are independent.
Bayes’ theorem
Bayes’ theorem allows us to squeeze an additional probability out of a certain set
of known probabilities. Before we state the theorem, let’s look at an example.
example 73
A grocery store receives an order from Supplier A that consists of 57% lettuce and
43% spinach. It also receives an order from Supplier B that consists of 39% lettuce
and 61% spinach. Before the orders are unloaded, a clerk randomly selects an order
box and pulls out a package of produce to show to the produce manager. Later the
grocer is notified that spinach from Supplier B is contaminated. If the clerk pulled out
a package of spinach, what is the probability that it came from Supplier B?
Let E 1 be the event that the package came from Supplier A and E 2 be the event
that the package came from Supplier B. Let F be the event that the package was
spinach. The sample space looks something like
{AL1, AL2, AL3, …, AS1, AS2,AS3, …, BL1, BL2,BL3, …, BS1, BS2, BS3 …}
E 1 and E 2 are disjoint events and E 1 c E 2 = S.
We know that
P(E 1 ) = 1/2
(equally likely that either box was chosen)
P(E 2 ) = 1/2
P(F 0 E 1 ) = 43/100
(percentage of spinach in Supplier A order)
P(F 0 E 2 ) = 61/100
(percentage of spinach in Supplier B order)
and we want
P(E 2 0 F )
(probability the package came from Supplier B
given that it was spinach)
Although we know that the probability of the clerk choosing the Supplier B order
box is 0.5, we suspect that P(E 2 0 F ) is greater than 0.5 because the item was spinach
and Supplier B has a higher percentage of spinach than Supplier A. It turns out that
we can compute this probability by sufficient fiddling with the probabilities we do
have.
From the definition of conditional probability,
P(E 2 0 F ) =
P(F d E 2 )
P(F)
or P(F d E 2 ) = P(E 2 0 F )P(F )
ReminDeR
P(E 1 c E 2 ) = P(E 1 )+P(E 2 )
only when E 1 and E 2 are
disjoint events. P(E 1 d E 2 )
= P(E 1 ) # P(E 2 ) only when
E 1 and E 2 are independent
events.
Sets, Combinatorics, and Probability
At this point we have what might be called an addition rule and a multiplication rule for probability, loosely related to the addition principle and the multiplication principle for counting. If we want the probability of event E 1 or event
E 2 , that is, P(E 1 c E 2 ), we can add the respective probabilities—but only if the
events are disjoint. If we want the probability of event E 1 and event E 2 , that is,
P(E 1 d E 2 ), we can multiply the respective probabilities—but only if the events
are independent.
Bayes’ theorem
Bayes’ theorem allows us to squeeze an additional probability out of a certain set
of known probabilities. Before we state the theorem, let’s look at an example.
example 73
A grocery store receives an order from Supplier A that consists of 57% lettuce and
43% spinach. It also receives an order from Supplier B that consists of 39% lettuce
and 61% spinach. Before the orders are unloaded, a clerk randomly selects an order
box and pulls out a package of produce to show to the produce manager. Later the
grocer is notified that spinach from Supplier B is contaminated. If the clerk pulled out
a package of spinach, what is the probability that it came from Supplier B?
Let E 1 be the event that the package came from Supplier A and E 2 be the event
that the package came from Supplier B. Let F be the event that the package was
spinach. The sample space looks something like
{AL1, AL2, AL3, …, AS1, AS2,AS3, …, BL1, BL2,BL3, …, BS1, BS2, BS3 …}
E 1 and E 2 are disjoint events and E 1 c E 2 = S.
We know that
P(E 1 ) = 1/2
(equally likely that either box was chosen)
P(E 2 ) = 1/2
P(F 0 E 1 ) = 43/100
(percentage of spinach in Supplier A order)
P(F 0 E 2 ) = 61/100
(percentage of spinach in Supplier B order)
and we want
P(E 2 0 F )
(probability the package came from Supplier B
given that it was spinach)
Although we know that the probability of the clerk choosing the Supplier B order
box is 0.5, we suspect that P(E 2 0 F ) is greater than 0.5 because the item was spinach
and Supplier B has a higher percentage of spinach than Supplier A. It turns out that
we can compute this probability by sufficient fiddling with the probabilities we do
have.
From the definition of conditional probability,
P(E 2 0 F ) =
P(F d E 2 )
P(F)
or P(F d E 2 ) = P(E 2 0 F )P(F )
ReminDeR
P(E 1 c E 2 ) = P(E 1 )+P(E 2 )
only when E 1 and E 2 are
disjoint events. P(E 1 d E 2 )
= P(E 1 ) # P(E 2 ) only when
E 1 and E 2 are independent
events.