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WARM-UP 9/4/12 Fifteen persons reporting to a Red Cross center one day are typed for blood, and the following counts are found: Blood group O A B AB Total No. of persons 3 5 6 1 15 If one person is randomly selected, what is the probability that this person’s blood group is: a) AB? b) Either A or B? c) Not O? ADDITION LAW P(AUB)=P(A)+P(B)-P(AB) Special Addition Law for Incompatible Events P(AUB)=P(A)+P(B) QUESTION: 52 • What is the probability of getting a card that is either a black card or and ace? DEMONSTRATION Conditional Probability The Basics QUESTION 1: • Suppose we have 14 marbles in a bag; 6 are green, 4 are red and 4 are yellow. What is the probability of pulling a red marble, given that the marble we just pulled is a primary color? 14 MARBLES • 6 Green, 4 Red, 4 Yellow 14 MARBLES • P(A)= probability of getting a red marble • P(B)= probability of getting a marble that’s a primary color P(B)=8/14 • 8 of the 14 marbles are primary colors P(AB)=4/14 • 4 of the 8 marbles are red P(A|B)=P(AB)/P(B)=4/8 • We were told marble was primary color, and only 4 of those marbles could possibly be red P(A|B)=P(AB)/P(B) • The probability of an event A, given that an event B has already occurred • How likely will A happen, since B has already happened? QUESTION: • A family has two children. What is the conditional probability that both are boys given that at least one of them is a boy? SAMPLE SPACE P(A)=getting at least one boy • 3 of 4 outcomes gives at least one boy P(A)=getting at least one boy • 1 out of the 3 outcomes gives us both boys P(A)=getting at least one boy P(B|A)=P(AB)/P(A)=1/3 BODY WEIGHT AND HYPERTENSION Overweight Normal Weight Underweight Total Hypertensive .10 .08 .02 .20 Not Hypertensive .15 .45 .20 .80 Total .25 .53 .22 1.00 a) What is the probability that a person selected at random from this group will have hypertension? b) A person selected at random from this group is found to be underweight. What is the probability that this person is also hypertensive? BODY WEIGHT AND HYPERTENSION Overweight Normal Weight Underweight Total Hypertensive .10 .08 .02 .20 Not Hypertensive .15 .45 .20 .80 Total .25 .53 .22 1.00 a) Because 20% of the group is hypertensive and the individual is selected at random from this group, we conclude that P(A)=.2. This is unconditional probability of A BODY WEIGHT AND HYPERTENSION Overweight Normal Weight Underweight Total Hypertensive .10 .08 .02 .20 Not Hypertensive .15 .45 .20 .80 Total .25 .53 .22 1.00 b) When given the info that the selected person is underweight, we ignore the first and second columns. The third column shows that among the subgroup of underweight people the proportion of having hypertension is .02/.22=.091. KEY FORMULAS CONDTIONAL PROBABILITY P(A|B)=P(AB)/P(B) MULTIPLICATION LAW OF PROBABILITY P(AB)=P(B)P(A|B) COMPREHENSION EXAMPLE • http://www.khanacademy.org/math/probabili ty/v/independent-events-1 Let’s see if you understand what it means for two events to be independent or dependent UNDERSTANDING INDEPENDENCE There are 25 pens in a container on your desk. Among them, 20 will write well but 5 have defective ink cartridges. You will select 2 pens to take a business appointment. What is the probability that both pens are defective? • Without replacement • With replacement INDEPENDENCE RELATION • Two events A and B are independent if P(A|B)=P(A) Equivalently P(B|A)=P(B) Or P(AB)=P(A)P(B) APPLICATION • http://www.khanacademy.org/math/probabili ty/v/independent-events-2 In this problem, we are using the formula: P(AB)=P(A|B)P(B) Since the two events are independent P(AB)=P(A)P(B), where P(A|B)=P(A) DRILL SET • Now, let’s practice what you have learned so far. You will have 2 to 3 minutes to solve each problem; I will be keeping track of time.