The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL STOR 155 Introductory Statistics Lecture 11: General Probability Rules 10/8/09 Lecture 11 1
Review Outcome, Sample space, Event Union (or), Intersection (and), Complement, Disjoint, Venn diagram Basic rules: For any event A, P( not A) = 1 - P(A). If A and B are disjoint, then P(A B) = 0. For any two events A and B, P(A B) = P(A) + P(B) - P(A B). 10/8/09 Lecture 11 2
General Addition Rule 10/8/09 Lecture 11 3
Independence A and B are independent if knowing that one occurs does not change the probability that the other occurs. For independent events A and B, P(A B) = P(A) P(B) (Multiplication rule) 10/8/09 Lecture 11 4
Cards (Ex 1) A card is drawn from a deck of 52 playing cards. What is the probability that the card is -- a club? (event A) -- a king? (event B) -- a club and a king? (event A B) Are A and B independent? 10/8/09 Lecture 11 5
Independent vs Disjoint A and B are independent if and only if P(A B) = P(A) P(B) If A and B are disjoint, P(A B) = 0 Note: If P(A) > 0 and P(B) > 0, then disjoint A, B are dependent. (A happens simply implies that B does not happen!) 10/8/09 Lecture 11 6
Conditional Probability The probability of an event measures how likely it will occur. A conditional probability predicts how likely an event will occur under specified conditions. P(A B) = the conditional probability that A occurs (uncertain), given that B has occurred (certain). The condition contains partial knowledge. P( A B) P( A P( B) B) 10/8/09 Lecture 11 7
Two-way Table (Ex 2) Prediction record of a TV weather forecaster over the past several years: Forecast Sunny cloudy Rainy Row Sum Sunny.50.05.04.59 Actual Cloudy.04.10.02.16 Weather Rainy.10.05.10.25 Column Sum.64.20.16 1 How likely was the forecaster wrong? What was the probability of rain? What was the probability of rain given the forecast was sunny? 10/8/09 Lecture 11 8
Gender of Children (Ex 3) A family has two children. Assume all four possible outcomes (younger is a boy, older is a girl; ) are equally likely. What is the probability that both are boys given that at least one is a boy? Let A = {both are boys}, B = {at least one is a boy}. P(A) = 1/4, P(B) = 3/4. (Why?) P(A B) = P(A B) / P(B) = P(A) / P(B) = 1/3. 10/8/09 Lecture 11 9
Stock Market (Ex 4) Q: The probability that a mutual fund company will get increased contributions from investors is? A: The following information is gathered the probability becomes 0.9 if the stock market goes up. the probability drops below 0.3 if the stock market drops. with probability 0.4, the stock market rises. So? The events of interest are: A = {the stock market rises} B = {the company receives increased contributions} Calculate P(A B) and P(B). 10/8/09 Lecture 11 10
Urn of Chips (Ex 5) An urn contains 5 white chips and 4 blue chips. Two chips are drawn sequentially without replacement. What is the probability of obtaining the sequence (white, blue)? A = {1st chip is white}, B = {2nd chip is blue}. Want to know P(A B) =? Use the formula P(A B) = P(B A) P(A) 10/8/09 Lecture 11 11
More Independence Conditions For any two events A and B, P(A B) = P(A B) P(B) = P(B A) P(A) Two events A and B are independent If any of the following equivalent conditions holds: (i) P(A B) = P(A) (ii) P(B A) = P(B) (iii) P(A B) = P(A) P(B) 10/8/09 Lecture 11 12
Cards (Ex 1 revisited) A card is drawn from a deck. What is the probability that the card is a club, given the card is a king? A = {the card is a king}, B = {the card is a club}. P(A) = 4/52, P(B A) = 1/52. Hence P(B A) = P(B A) / P(A) = 1/4. Note that P(B A)= P(B). This means knowing A occurs has no impact on the chance for occurrence of B. In other words, B & A are independent. 10/8/09 Lecture 11 13
High school athlete (Ex 6) 5% of male high school athletes go on to play at college level. Of these, 1.7% enter major league professional sports. About 0.01% of the high school athletes who never compete in college enter professional sports. A = {competes in college} B = {competes professionally} Q: the probability that a high school athlete competes in college and then goes on to have a pro career? Q: a high school athlete goes to professional sports? 10/8/09 Lecture 11 14
Tree Diagram (Ex 6 continued) 10/8/09 Lecture 11 15
Bayes Rule For any two events A and B, P(A B) = P(A B) P(B) P(B) = P(B A) + P(B { not A} ) = P(B A) P(A) + P(B not A) P( not A) If P(A) and P(B) are not 0 or 1, then P(A B) = P(A B) / P(B) = 10/8/09 Lecture 11 16
Quality Control (Ex 7) A manufacturer is trying to find ways to reduce # of defective parts. The present procedure produces 5% defectives. An inspection found that 40% of defectives and 15% of non-defectives were produced by machine 1. Q: What is the probability that a part produced by machine 1 is found to be defective? --- Try a tree diagram. 10/8/09 Lecture 11 17
Quality Control (Ex 7 continued) A = {a part is defective} B = {a part is produced by machine 1} P(A B) =? P(A) =.05, P( not A) =.95, P(B A) =.4, and P(B not A) =.15..4.05 P( A B).4.05.15.95 12.3%. 10/8/09 Lecture 11 18
Lab Test (Ex 8) A lab test yields two possible results: positive or negative. 99% of people with a particular disease will produce a positive result. But 2% of people without the disease will also produce a positive result. Suppose that 0.1% of the population actually has the disease. Q: What is the probability that a person chosen at random actually has the disease, given a positive result? --- Try a tree diagram. 10/8/09 Lecture 11 19
Lab Test (Ex 8 continued) D = {disease}, + = {positive test result}. Want P(D +) =? P(D) =.001, P( not D ) = 0.999. P(+ D) =.99, and P(+ not D ) =.02. Applying Bayes Rule,.99.001.00099 P( D ).99.001.02.999.02097 4.7%. 10/8/09 Lecture 11 20
Bayes Rule (general) 10/8/09 Lecture 11 21
Take Home Message General Addition Rule Independence, Conditional probability, Multiplication Rule Bayes Rule Tree Diagram 10/8/09 Lecture 11 22