# Chapter 14 Probability Basics Laws of Probability Odds and Probability Probability Trees.

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Chapter 14 Probability Basics Laws of Probability Odds and Probability Probability Trees

Birthday Problem What is the smallest number of people you need in a group so that the probability of 2 or more people having the same birthday is greater than 1/2? Answer: 23 No. of people23304060 Probability.507.706.891.994

Probability Formal study of uncertainty The engine that drives Statistics Primary objective of lecture unit 4: 1.use the rules of probability to calculate appropriate measures of uncertainty. 2.Learn the probability basics so that we can do Statistical Inference

Introduction Nothing in life is certain We gauge the chances of successful outcomes in business, medicine, weather, and other everyday situations such as the lottery or the birthday problem Tomorrow's Weather

A phenomenon is random if individual outcomes are uncertain, but there is nonetheless a regular distribution of outcomes in a large number of repetitions. Randomness and probability Randomness ≠ chaos

Coin toss The result of any single coin toss is random. But the result over many tosses is predictable, as long as the trials are independent (i.e., the outcome of a new coin flip is not influenced by the result of the previous flip). First series of tosses Second series The probability of heads is 0.5 = the proportion of times you get heads in many repeated trials.

4.1 The Laws of Probability 1.Relative frequency event probability = x/n, where x=# of occurrences of event of interest, n=total # of observations –Coin, die tossing; nuclear power plants? Limitations repeated observations not practical Approaches to Probability

Approaches to Probability (cont.) 2.Subjective probability individual assigns prob. based on personal experience, anecdotal evidence, etc. 3.Classical approach every possible outcome has equal probability (more later)

Basic Definitions Experiment: act or process that leads to a single outcome that cannot be predicted with certainty Examples: 1.Toss a coin 2.Draw 1 card from a standard deck of cards 3.Arrival time of flight from Atlanta to RDU

Basic Definitions (cont.) Sample space: all possible outcomes of an experiment. Denoted by S Event: any subset of the sample space S; typically denoted A, B, C, etc. Null event: the empty set  Certain event: S

Examples 1.Toss a coin once S = {H, T}; A = {H}, B = {T} 2.Toss a die once; count dots on upper face S = {1, 2, 3, 4, 5, 6} A=even # of dots on upper face={2, 4, 6} B=3 or fewer dots on upper face={1, 2, 3} 3.Select 1 card from a deck of 52 cards. S = {all 52 cards}

Laws of Probability

Coin Toss Example: S = {Head, Tail} Probability of heads = 0.5 Probability of tails = 0.5 3 ) The complement of any event A is the event that A does not occur, written as A. The complement rule states that the probability of an event not occurring is 1 minus the probability that is does occur. P(not A) = P(A) = 1 − P(A) Tail = not Tail = Head P(Tail ) = 1 − P(Tail) = 0.5 Probability rules (cont’d) Venn diagram: Sample space made up of an event A and its complement A, i.e., everything that is not A.

Birthday Problem What is the smallest number of people you need in a group so that the probability of 2 or more people having the same birthday is greater than 1/2? Answer: 23 No. of people23304060 Probability.507.706.891.994

Example: Birthday Problem A={at least 2 people in the group have a common birthday} A’ = {no one has common birthday}

Unions: , or Intersections: , and A  A 

Mutually Exclusive (Disjoint) Events Mutually exclusive or disjoint events-no outcomes from S in common A and B disjoint: A  B=  A and B not disjoint A  A  Venn Diagrams

Addition Rule for Disjoint Events 4. If A and B are disjoint events, then P(A or B) = P(A) + P(B)

Laws of Probability (cont.) General Addition Rule 5. For any two events A and B P(A or B) = P(A) + P(B) – P(A and B)

20 For any two events A and B P(A or B) = P(A) + P(B) - P(A and B) A B P(A) =6/13 P(B) =5/13 P(A and B) =3/13 A or B + _ P(A or B) = 8/13 General Addition Rule

Laws of Probability (cont.) Multiplication Rule 6. For two independent events A and B P(A and B) = P(A) × P(B) Note: assuming events are independent doesn’t make it true.

Multiplication Rule The probability that you encounter a green light at the corner of Dan Allen and Hillsborough is 0.35, a yellow light 0.04, and a red light 0.61. What is the probability that you encounter a red light on both Monday and Tuesday? It’s reasonable to assume that the color of the light you encounter on Monday is independent of the color on Tuesday. So P(red on Monday and red on Tuesday) = P(red on Monday) × P(red on Tuesday) = 0.61 × 0.61 = 0.3721

Laws of Probability: Summary 1. 0  P(A)  1 for any event A 2. P(  ) = 0, P(S) = 1 3. P(A’) = 1 – P(A) 4. If A and B are disjoint events, then P(A or B) = P(A) + P(B) 5. For any two events A and B, P(A or B) = P(A) + P(B) – P(A and B) 6. For two independent events A and B P(A and B) = P(A) × P(B)

M&M candies ColorBrownRedYellowGreenOrangeBlue Probability0.30.2 0.1 ? If you draw an M&M candy at random from a bag, the candy will have one of six colors. The probability of drawing each color depends on the proportions manufactured, as described here : What is the probability that an M&M chosen at random is blue? What is the probability that a random M&M is any of red, yellow, or orange? S = {brown, red, yellow, green, orange, blue} P(S) = P(brown) + P(red) + P(yellow) + P(green) + P(orange) + P(blue) = 1 P(blue) = 1 – [P(brown) + P(red) + P(yellow) + P(green) + P(orange)] = 1 – [0.3 + 0.2 + 0.2 + 0.1 + 0.1] = 0.1 P(red or yellow or orange) = P(red) + P(yellow) + P(orange) = 0.2 + 0.2 + 0.1 = 0.5

Example: toss a fair die once S = {1, 2, 3, 4, 5, 6} A = even # appears = {2, 4, 6} B = 3 or fewer = {1, 2, 3} P(A or  B) = P(A) + P(B) - P(A and  B) =P({2, 4, 6}) + P({1, 2, 3}) - P({2}) = 3/6 + 3/6 - 1/6 = 5/6

Chapter 14 (cont.) zOdds and Probabilities zProbability Trees

ODDS AND PROBABILITIES zWorld Series OddsWorld Series Odds zFrom probability to odds zFrom odds to probability

From Probability to Odds zIf event A has probability P(A), then the odds in favor of A are P(A) to 1-P(A). It follows that the odds against A are 1-P(A) to P(A) z If the probability of an earthquake in California is.25, then the odds in favor of an earthquake are.25 to.75 or 1 to 3. The odds against an earthquake are.75 to.25 or 3 to 1

From Odds to Probability zIf the odds in favor of an event E are a to b, then P(E)=a/(a+b) zin addition, P(E’)=b/(a+b) z If the odds in favor of UNC winning the NCAA’s are 3 (a) to 1 (b), then P(UNC wins)=3/4 P(UNC does not win)= 1/4

Chapter 14 (cont.) Probability Trees A Graphical Method for Complicated Probability Problems

Example: Prob. of playing pro baseball  Data shows the following probabilities concerning high school baseball players: 6.1% of hs players go on to play at the college level 9.4% of hs players that play in college go on to play professionally 0.2% of hs players that do not compete in college play pro baseball  What is the probability that a high school baseball ultimately plays professional baseball?

Probability Tree Approach zA probability tree is a useful way to visualize this problem and to find the desired probability.

Q1: probability that a high school baseball ultimately plays professional baseball? P(play college and play professionally) =.061*.094 =.005734 P(do not play college and play professionally) =.939*.002 =.001878 P(high school baseball player plays professional baseball) =.005734 +.001878 =.007612

Question 2 zGiven that a high school player played professionally, what is the probability he played in college? P(play college and play professionally) =.061*.094 =.005734 P(do not play college and play professionally) =.939*.002 =.001878

Example: AIDS Testing zV={person has HIV}; CDC: Pr(V)=.006 zP : test outcome is positive (test indicates HIV present) zN : test outcome is negative zclinical reliabilities for a new HIV test: 1.If a person has the virus, the test result will be positive with probability.999 2.If a person does not have the virus, the test result will be negative with probability.990

Question 1 zWhat is the probability that a randomly selected person will test positive?

Probability Tree Multiply branch probs clinical reliability

Question 1 Answer zWhat is the probability that a randomly selected person will test positive? zPr(P )=.00599 +.00994 =.01593

Question 2 zIf your test comes back positive, what is the probability that you have HIV? (Remember: we know that if a person has the virus, the test result will be positive with probability.999; if a person does not have the virus, the test result will be negative with probability.990). zLooks very reliable

Question 2 Answer Answer two sequences of branches lead to positive test; only 1 sequence represented people who have HIV. Pr(person has HIV given that test is positive) =.00599/(.00599+.00994) =.376

Summary zQuestion 1: zPr(P ) =.00599 +.00994 =.01593 zQuestion 2: two sequences of branches lead to positive test; only 1 sequence represented people who have HIV. Pr(person has HIV given that test is positive) =.00599/(.00599+.00994) =.376

Recap zWe have a test with very high clinical reliabilities: 1.If a person has the virus, the test result will be positive with probability.999 2.If a person does not have the virus, the test result will be negative with probability.990 zBut we have extremely poor performance when the test is positive: Pr(person has HIV given that test is positive) =.376 zIn other words, 62.4% of the positives are false positives! Why? zWhen the characteristic the test is looking for is rare, most positives will be false.

examples 1. P(A)=.3, P(B)=.4; if A and B are mutually exclusive events, then P(A  B)=? A  B = , P(A  B) = 0 2. 15 entries in pie baking contest at state fair. Judge must determine 1 st, 2 nd, 3 rd place winners. How many ways can judge make the awards? 15 P 3 = 2730

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