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Warm-up Define the sample space of each of the following situations… Define the sample space of each of the following situations… Choose a student in your class at random. Ask how much time that student spent studying during the past 24 hours. Choose a student in your class at random. Ask how much time that student spent studying during the past 24 hours. The Physician’s Health Study asked 11,000 physicians to take an aspirin every other day and observed how many of them had a heart attack in a five-year period. The Physician’s Health Study asked 11,000 physicians to take an aspirin every other day and observed how many of them had a heart attack in a five-year period. In a test of new package design, you drop a carton of a dozen eggs from a height of 1 foot and count the number of broken eggs. In a test of new package design, you drop a carton of a dozen eggs from a height of 1 foot and count the number of broken eggs.

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Section 6.2 Section 6.2 Independence and the Multiplication Rule

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Two Special Rules We’ve learned the addition rule for disjoint events: If A and B are disjoint, then P(A or B) = P(A) + P(B). We’ve learned the addition rule for disjoint events: If A and B are disjoint, then P(A or B) = P(A) + P(B). Now we’ll learn the multiplication rule for independent events: that if A and B are independent, then P(A and B) = P(A)●P(B) Now we’ll learn the multiplication rule for independent events: that if A and B are independent, then P(A and B) = P(A)●P(B) Remember that two events are independent if knowing that one occurs does not change the probability that the other occurs. Remember that two events are independent if knowing that one occurs does not change the probability that the other occurs.

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Examples of Independent Events Toss a coin twice. Let A = first toss is a head and B = second toss is a head. Events A and B are independent. Thus, the P(A ∩B) = P(A)*P(B). Toss a coin twice. Let A = first toss is a head and B = second toss is a head. Events A and B are independent. Thus, the P(A ∩B) = P(A)*P(B). P(Head and Tail) = P(Head) * P(Tail) = ½ * ½ = ¼ P(Head and Tail) = P(Head) * P(Tail) = ½ * ½ = ¼ Draw 3 cards from a deck, replacing and shuffling in between each draw. This is called “with replacement.” Draw 3 cards from a deck, replacing and shuffling in between each draw. This is called “with replacement.”

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Without Replacement If you draw three cards from a deck without replacing, the probabilities change on each draw. Therefore, drawing without replacement is NOT independent. If you draw three cards from a deck without replacing, the probabilities change on each draw. Therefore, drawing without replacement is NOT independent.

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Cautions The addition rule for disjoint events and the multiplication rule for independent events only work when the criteria are met. Resist the temptation to use them for events that are not disjoint or not independent. The addition rule for disjoint events and the multiplication rule for independent events only work when the criteria are met. Resist the temptation to use them for events that are not disjoint or not independent. You must be told or have prior knowledge that an event is disjoint or independent. You must be told or have prior knowledge that an event is disjoint or independent. Do not confuse disjoint with independent. Disjoint can be displayed in a Venn Diagram. Independence can not. Do not confuse disjoint with independent. Disjoint can be displayed in a Venn Diagram. Independence can not.

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Sample Questions Suppose that among the 6000 students at a high school, 1500 are taking honors courses and 1800 prefer watching basketball to watching football. If taking honors courses and preferring basketball are independent, how many students are both taking honors courses and prefer basketball to football? Suppose that among the 6000 students at a high school, 1500 are taking honors courses and 1800 prefer watching basketball to watching football. If taking honors courses and preferring basketball are independent, how many students are both taking honors courses and prefer basketball to football?

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Sample Questions Suppose that for any given year, the probabilities that the stock market declines is.4, and the probability that women’s hemlines are lower is.35. Suppose that the probability that both events occur is.3. Are the two events independent? Suppose that for any given year, the probabilities that the stock market declines is.4, and the probability that women’s hemlines are lower is.35. Suppose that the probability that both events occur is.3. Are the two events independent?

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Sample Questions In a 1974 “Dear Abby” letter, a woman lamented that she had just given birth to her eighth child, and all were girls! Her doctor had assured her that the chance of the eighth child being a girl was only 1 in 100. In a 1974 “Dear Abby” letter, a woman lamented that she had just given birth to her eighth child, and all were girls! Her doctor had assured her that the chance of the eighth child being a girl was only 1 in 100. A) What was the real probability that the eighth child would be a girl? A) What was the real probability that the eighth child would be a girl? B) Before the birth of the first child, what was the probability that the woman would give birth to eight girls in a row? B) Before the birth of the first child, what was the probability that the woman would give birth to eight girls in a row?

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Section 6.3 General Probability Rules

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Special Probability Rules If A and B are disjoint, then P(AUB) = P(A) + P(B). If A and B are disjoint, then P(AUB) = P(A) + P(B). This rule relies on a condition to be met. This rule relies on a condition to be met. So, what if the events are NOT disjoint? So, what if the events are NOT disjoint? What if there are more than 2 disjoint events? What if there are more than 2 disjoint events?

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General Probability Rules P(AUB) = P(A) + P(B) – P(A ∩ B) P(AUB) = P(A) + P(B) – P(A ∩ B) This is how the formula will appear on your formula sheet for the exam (and thus my tests). This is how the formula will appear on your formula sheet for the exam (and thus my tests). If A and B are disjoint, then what does P(A∩B) equal?

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Sample Question: Student Survey (making an A in Statistics only) = 18 students (making an A in Calculus only) = 63 students (making an A in Stats and in Calculus) = 27 students (making an A in Statistics only) = 18 students (making an A in Calculus only) = 63 students (making an A in Stats and in Calculus) = 27 students S=150 students S=150 students Draw a Venn diagram. Draw a Venn diagram. Find the probability of making an A in Stats but not in Calc. Find the probability of making an A in Stats but not in Calc. Find the probability of making an A in Calc but not in Stats. Find the probability of making an A in Calc but not in Stats. Find the probability of not making an A in either subject. Find the probability of not making an A in either subject. Find the probability of making an A in either Calc or Stats. Find the probability of making an A in either Calc or Stats. Are these events independent? Are these events independent?

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Extending the Rules to Three Events If you have three disjoint events, then P(A or B or C) = P(A) + P(B) + P(C) If you have three disjoint events, then P(A or B or C) = P(A) + P(B) + P(C) If they are not disjoint, then P(A or B or C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(A ∩ C) – P(B ∩ C) If they are not disjoint, then P(A or B or C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(A ∩ C) – P(B ∩ C) Often it is easier to draw a Venn Diagram than to use the formula. Often it is easier to draw a Venn Diagram than to use the formula. Look at p Look at p

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Sample Questions Suppose that the probability that you will receive an A in AP Statistics is.35, the probability that you will receive As in both AP Statistics and AP Biology is.19, and the probability that you will receive an A in AP Bio but not in Stats is.17. Which is a proper conclusion? Suppose that the probability that you will receive an A in AP Statistics is.35, the probability that you will receive As in both AP Statistics and AP Biology is.19, and the probability that you will receive an A in AP Bio but not in Stats is.17. Which is a proper conclusion? A) P(A in AP Bio) =.36 A) P(A in AP Bio) =.36 B) P(you didn’t take Bio) =.01 B) P(you didn’t take Bio) =.01 C) P(not making an A in AP Stat or Bio) =.38 C) P(not making an A in AP Stat or Bio) =.38 D) The given probabilities are impossible. D) The given probabilities are impossible.

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Sample Questions If P(A) =.2 and P(B) =.1, what is P(AUB) if A and B are independent? If P(A) =.2 and P(B) =.1, what is P(AUB) if A and B are independent? A).02 A).02 B).28 B).28 C).30 C).30 D).32 D).32 E) There is insufficient information to answer this question. E) There is insufficient information to answer this question.

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Sample Questions Given the probabilities P(A) =.4 and P(AUB) =.6, what is the probability P(B) if A and B are mutually exclusive? If A and B are independent? Given the probabilities P(A) =.4 and P(AUB) =.6, what is the probability P(B) if A and B are mutually exclusive? If A and B are independent? A).2,.28 A).2,.28 B).2,.33 B).2,.33 C).33,.2 C).33,.2 D).6,.33 D).6,.33 E).28,.2 E).28,.2

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Conditional Probability When events are not independent, the probability of one changes if we know that the other event occurred. When events are not independent, the probability of one changes if we know that the other event occurred. I will draw two cards without replacement. I will draw two cards without replacement. Let A = {1 st card I draw is an Ace} Let A = {1 st card I draw is an Ace} Let B = {2 nd card I draw is an Ace} Let B = {2 nd card I draw is an Ace} The probability of B occurring changes depending on whether A occurred. The probability of B occurring changes depending on whether A occurred. The new notation P(B|A) is read “the probability of B given A.” It asks you to find the probability of B knowing that A has occurred. The new notation P(B|A) is read “the probability of B given A.” It asks you to find the probability of B knowing that A has occurred.

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Let’s look at education and age Education Age 25 – – Total No high school 4,4749,15514,22427,853 Completed HS 11,54626,48120,06058,087 1 to 3 years college 10,70022,61811,12744, years of college 11,06623,18310,59644,845 Total37,78681,43556,008175,230

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Find these probabilities Let A = {the person chosen is 25 – 34} Let A = {the person chosen is 25 – 34} Let B = {the person chosen has 4+ years of college} Let B = {the person chosen has 4+ years of college} Find P(A). Find P(A). Find P(A and B). Find P(A and B). Find P(B|A). Find P(B|A). Given that a person has only a HS diploma, what is the probability that the person is 55 or older? Given that a person has only a HS diploma, what is the probability that the person is 55 or older?

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The general rule for multiplication P(A ∩ B) = P(A)*P(B|A) P(A ∩ B) = P(A)*P(B|A) This can be rewritten as This can be rewritten as P(B|A) = P(A ∩ B)/P(A) Example: Two cards are drawn without replacement. Example: Two cards are drawn without replacement. Let A = {1 st card is a spade} Let A = {1 st card is a spade} Let B = {2 nd card is a spade} Let B = {2 nd card is a spade} Find P(B|A). Find P(B|A). Find P(A and B). Find P(A and B).

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Next Example Suppose the probability that the dollar falls in value compared to the yen is 0.5. Suppose the probability that the dollar falls in value compared to the yen is 0.5. Suppose that the probability that if the dollar falls in value, a supplier from Japan will renegotiate their contract is 0.7. Suppose that the probability that if the dollar falls in value, a supplier from Japan will renegotiate their contract is 0.7. What is the probability that the dollar will fall in value and the supplier demands renegotiation? What is the probability that the dollar will fall in value and the supplier demands renegotiation?

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Homework Chapter 5# 4, 9, 40, 43, 51, 60, 83, 89, 94

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