Section 6.6: Some General Probability Rules

General Addition Rule for Two Events For any two events E and F,

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) =.60P(F) =.40P(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) = =.75

General Multiplication Rule For any two events E and F

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) =.20 P(E│F) =.80 P(E and F) = P(E│F)P(F) =(.80)(.20) =.16 Thus 16% of all teenager drivers in this country received a citation and attended traffic school.

Law of Total Probability If B 1 and B 2 are disjoint events with P(B 1 )+P(B 2 ) = 1, then for any event E, P(E) = P(E ∩ B 1 ) + P(E ∩ B 2 ) = P(E│B 1 )P(B 1 ) + P(E│B 2 )P(B 2 ) More generally, if B 1, B 2,…,B k are disjoint events with P(B 1 ) + P(B 2 )+…+P(B k ) = 1, then for any event E, P(E) = P(E ∩ B 1 ) + P(E ∩ B 2 ) +…+ P(E ∩ B k ) = P(E│B 1 )P(B 1 )+P(E│B 2 )P(B 2 )+…+P(E│B k )P(B k )

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) = =.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(B 1 )+P(B 2 ) = 1, then for any event E,

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: S 1 = event that package was shipped using Service 1 S 2 = event that package was shipped using Service 2 L = event that the package is late

The following probabilities are known: P(S 1 ) =.3 P(S 2 ) =.7 P(L│S 1 ) =.1 P(L│S 2 ) =.08 Because you know that your package is late you should use Baye’s Rule.

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.