1. 6.2 – Probability Models It is often important and necessary to provide a mathematical description or model for randomness.

2. Basic Definitions Probability: long-run proportion of repetitions on which an event can occur Sample Space: set of all possible outcomes  Use charts/lists/tree diagrams to display Multiplication Principle: if event 1 can occur in n ways and event 2 can occur in m ways, then they both can be done in n x m ways

3. Definitions continued… Probability model: mathematical description of a random phenomenon consisting of two parts 1.A sample space S 2.A way of assigning probabilities to events Being able to properly enumerate the outcomes in a sample space is critical to determining probabilities. Sometimes it's helpful to use a tree diagram.

4. Sample Space for the Sum of 2 Dice There are 36 total possible ways to roll 2 dice. The sample space contains 11 outcomes: S = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

5. What is the sample space when we toss a coin and roll a die?

6. Using the Multiplication Principle Since die 1 can occur in 6 ways, and die 2 can occur in 6 ways, they can occur in 6 x 6 = 36 ways together How many possible outcomes are there for tossing a coin 4 times?  Each toss has 2 possibilities  2 x 2 x 2 x 2 = 16 possible outcomes

7. Nondiscrete Sample Space Some sample spaces are too large to allow all of the possible outcomes to be listed Ex: Many computing systems have a function that will generate a random number between 0 and 1. The sample space is S = {all numbers between 0 and 1) Due to limited decimal places, not all numbers are possible CW: pg. 410 #6.22, 6.24 HW: pg. 416 #6.29-6.36

8. Probability Rules If A is an event, then the probability P(A) is a number between 0 and 1, inclusive. Prob(A does not happen) = P(A C ) = 1 – P(A)  This is also called the complement of event A If S is the sample space in a probability model, then P(S) = 1

9. Probability Rules Continued Two events are disjoint if they have no outcomes in common.  If A and B are disjoint, then P(A or B ) = P(A) + P(B)  Some sources use the phrase mutually exclusive to describe disjoint events.  Roll two dice, If A = “roll sum of 7” and B = “roll a sum of 2”, then P(A or B) = 6/36 + 1/36 = 7/36 since A and B are disjoint  If C = “roll a sum of 7” and D = “roll and odd sum” then C and D are not disjoint since 7 is an odd sum

10. Probability Rules Continued Two events are independent if knowing that one occurs does not change the probability of the other.  If A and B are independent, then P(A & B ) = P(A)* P(B)  If A = “ getting a head on the first coin toss” and B = “getting a head on the second coin toss,” then A and B are independent.  P(A & B) = (1/2)(1/2) = 1/4  If you are dealing two cards, but do not replace the first before the second is dealt, they are not independent.

11. Using Set Notation A U B means A union B  The set of all outcomes that are either in A or in B  Another way to indicate A or B A B mean A intersect B  The set of all outcomes that are in A and B  Another way to indicate A and B  If A and B are disjoint then A B = Ø

12. Venn Diagrams can also help Mutually exclusive events Showing the event A B

13. Probabilities for Rolling Dice When rolling two dice, what is the probability of rolling a sum of 5?

14. Assigning Probabilities: Finite Number of Outcomes

15. Benford’s Law It is a striking fact that the first digits of numbers in legitimate records often follow a distribution known as Benford’s Law. Note the first digit cannot be zero This law can help deter many crooks from faking tax returns, payment records, etc.

16. Using Benford’s Law Consider the events A = {1 st digit is 1} B = {1 st digit is 6 or greater} C = {1 st digit is odd} Find the following probabilities 1) P(A) = P(1) = 0.301 2) P(B) = P(6) + P(7) + P(8) + P(9) = 0.067 + 0.058 + 0.051 + 0.046 = 0.222 3) P(A C ) = 1 – P(A) = 1 – 0.301 = 0.699

17. Using Benford’s Law continued 4) P(A or B) = P(A) + P(B)(disjoint) = 0.301 + 0.222 = 0.523 5) P(C) = P(1) + P(3) + P(5) + P(7) + P(9) = 0.609 6) P(B or C) = P(1) + P(3) + P(5) + P(6) + P(7) + P(8) + P(9) = 0.727 Not P(B) + P(C) because they are not disjoint

18. Equally Likely Outcomes You might have thought that the first digits are distributed “at random” among digits 1 to 9. Because the total probability must be 1, the probability of each 9 outcomes must be 1/9 A crook who fakes data by using “random” digits will end up with too many first digits 6 or greater and too few 1’s and 2’s.

19. Equally Likely Outcomes Most random phenomenon do not have equally likely outcomes, so the general rule for finite sample spaces is more important than the special rule for equally likely outcomes. HW: pg. 423 #6.37, 6.38, 6.40, 6.41, 6.44

20. Independence and the Multiplication Rule The addition rule (for disjoint events) described the probability that one or the other of the events A and B will occur We will describe the probabilities that both events A and B occurs  Only in special situations  More details on the rule in 6.3

21. Toss a coin twice. A = 1 st toss is a head B = 2 nd toss is a head A and B are not disjoint. They can occur together whenever both tosses give heads. P(A and B)  Since they are independent, P(A and B) = P(A)P(B) = (½)(½) = ¼

22. Note: If two events are disjoint, then they cannot be independent. Knowing that one happened means that the other one cannot.

23. AIDS Testing One rapid HIV test has a probability of about 0.004 of producing a false-positive If the clinic tests 200 people, what is the chance of getting at least one false-positive?  P(negative result) = 1 – 0.004 = 0.996  P(at least one positive) = 1 – P(no positives) = 1 - P(200 negatives) = 1 –.996 200 = 1 - 0.4486 =.5514 The probability is greater than ½ that at least 1 of the 200 people will test positive for HIV even though no one has the virus