1. Probability of Equally Likely Outcomes 2. Complement Rule 1

Let S be a sample space consisting of N equally likely outcomes. Let E be any event. Then 2

 Suppose that a cruise ship returns to the US from the Far East. Unknown to anyone, 4 of its 600 passengers have contracted a rare disease. Suppose that the Public Health Service screens 20 passengers, selected at random, to see whether the disease is present aboard ship. What is the probability that the presence of the disease will escape detection? 3

 The sample space consists of samples of 20 drawn from among the 600 passengers. 4 There are = 596 non infected passengers. The disease is not detected if the 20 passengers are chosen from this group. Let E be the event that the disease is not detected.

 The number of outcomes in E is 5

 Complement Rule Let E be any event, E ' its complement. Then  Pr( E ) = 1 - Pr( E '). 6

 A group of 5 people is to be selected at random. What is the probability that 2 or more of them have the same birthday?  For simplicity we will ignore leap years and assume that each of the 365 days of the year are equally likely.  Choosing a person at random is equivalent to choosing a birthday at random. 7

 There are ways of choosing 5 birthdays.  It will also be easier to first find the probability that the 5 birthdays are different. Let this be E '.  The number of outcomes where the 5 birthdays are different is 365  364  363  362 

 Pr( E ) = 1- Pr( E ') = =.027  This example could be worked for a general group of r people.  In this case Probability that, in a randomly selected group of r people, at least two will have the same birthday r Pr(E)

 For a sample space with a finite number of equally likely outcomes, the probability of an event is the number of elements in the event divided by the number of elements in the sample space.  The probability of the complement of an event is 1 minus the probability of the event. 10