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UNC-Chapel Hill STOR 155 - Lecture 11 - General Probability Rules

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2/24/11 Lecture 11 1 STOR 155 Introductory Statistics Lecture 11: General Probability Rules The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL2/24/11 Lecture 11 2 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).2/24/11 Lecture 11 3 General Addition Rule2/24/11 Lecture 11 4 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)2/24/11 Lecture 11 5 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?2/24/11 Lecture 11 6 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!)2/24/11 Lecture 11 7 • 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. )()()|(BPBAPBAPConditional Probability2/24/11 Lecture 11 8 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”?2/24/11 Lecture 11 9 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) = 1/3.2/24/11 Lecture 11 10 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). Stock Market (Ex 4)2/24/11 Lecture 11 11 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)2/24/11 Lecture 11 12 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)2/24/11 Lecture 11 13 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.2/24/11 Lecture 11 14 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?2/24/11 Lecture 11 15 Tree Diagram (Ex 6 continued)2/24/11 Lecture 11 16 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) = …2/24/11 Lecture 11 17 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.2/24/11 Lecture 11 18 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. %.3.12.95.15.05.4.05.4 )|( BAP2/24/11 Lecture 11 19 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.2/24/11 Lecture 11 20 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, %.7.402097.00099..999.02.001.99.001.99 )|( DP2/24/11 Lecture 11


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UNC-Chapel Hill STOR 155 - Lecture 11 - General Probability Rules

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