Section 6.6: Some General Probability Rules

General Addition Rule for Two Events

• For any two events E and F,

)()()()( FEPFPEPFEP

Example

• Suppose that 60% of all customers of a large insurance agency have automobile policies with the agency, 40% have homeowner’s policies, and 25% have both types of policies. If a customer is randomly selected, what is the probability that he or she has at least one of these two types of policies with the agency?

• Let:

E = the event that a selected customer has auto insurance with the agency

F = the event that a selected customer has homeowner’s insurance with the agency

The given information implies that

P(E) = .60 P(F) = .40 P(E∩F) = .25

• We can obtain:P(customer has at least one of the two types

of policy)

= P(E ∪ F) = P(E) + P(F) – P(E ∩ F) = .60 + .40 - .25= .75

General Multiplication Rule

• For any two events E and F

)()()( FPFEPFEP

Example

• Suppose that 20% of all teenage drivers in a certain county received a citation for a moving violation within the past year. Assume in addition that 80% of those receiving such a citation attended traffic school so that the citation would not appear on their permanent driving record. If a teenage driver from this country is randomly selected, what is the probability that he or she received a citation and attended traffic school?

• Let’s define two events E and F as follows:

E = selected driver attended traffic school

F = selected driver received such a citation

P(F) = .20P(E│F) = .80

P(E and F) = P(E│F)P(F)

=(.80)(.20)=.16Thus 16% of all teenager drivers in this country

received a citation and attended traffic school.

Law of Total Probability

If B1 and B2 are disjoint events with P(B1)+P(B2) = 1, then for any event E,

P(E) = P(E ∩ B1) + P(E ∩ B2)

= P(E│B1)P(B1) + P(E│B2)P(B2)

More generally, if B1, B2,…,Bk are disjoint events with P(B1) + P(B2)+…+P(Bk) = 1, then for any event E,

P(E) = P(E ∩ B1) + P(E ∩ B2) +…+ P(E ∩ Bk)

= P(E│B1)P(B1)+P(E│B2)P(B2)+…+P(E│Bk)P(Bk)

Example

• An article gave information on bicycle helmet usage in some Cleveland suburbs. In Beachwood a safety education program and a helmet law were in place, whereas in Morland Hills neither a helmet law nor a safety education program is in place. The article reported that 68% of elementary school students from the city that has the rules always wear a helmet when bicycling, but only 21% of the students from Morland Hills reported that they always wear a helmet.

• Let define:

B = selected student is from Beachwood

M = selected student is from Morland Hills

H = selected student reports that he or she always wears a helmet when bicycling

• We can reason that

P(B) = P(M) = .5

P(H│B) = .68

P(H│M) = .21

What proportion of elementary school students in these two communities always wear helmets?

P(H) = P(H│B)P(B) + P(H│M)P(M)

= (.68)(.5) + (.21)(.5)

= .34 + .105

= .445

That is 44.5% of the elementary school children in these two communities always wear a helmet when cycling.

Bayes’ Rule

If B1 and B2 are disjoint events with P(B1)+P(B2) = 1, then for any event E,

)()()()(

)()()(

2211

111 BPBEPBPBEP

BPBEPEBP

Example

• Two shipping services offer overnight delivery of parcels, and both promise delivery before 10 AM. A mail-order catalog company ships 30% of its overnight packages using Shipping Service 1 and 70% using Service 2. Service 1 fails to meet the 10 AM delivery promise 10% of the time, whereas Service 2 fails to deliver by 10 AM 8% of the time. Suppose that you made a purchase from this company and were expecting your package by 10 AM, but it is late. Which shipping service is more likely to have been used?

• Let’s define the following events:

S1 = event that package was shipped using Service 1

S2 = event that package was shipped using Service 2

L = event that the package is late

• The following probabilities are known:

P(S1) = .3

P(S2) = .7

P(L│S1) = .1

P(L│S2) = .08

Because you know that your package is late you should use Baye’s Rule.

6512.086.

056.

)7)(.08(.)3)(.1(.

)7)(.08(.)(

3488.086.

03.

)7)(.08(.)3)(.1(.

)3)(.1(.)(

2

1

LSP

LSP

• So you should call service 2 to find your package.

• Even though they have a smaller percentage of late packages, it is more likely that a package was sent late because they ship out more packages.