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Review Conditional Probability –Denoted P(A|B), it represents the probability that A will occur given that B occurred. Independent events –Two events A and B are independent if: the occurrence of one does not affect the probability of the other. P(A|B)=P(A) P(B|A)=P(B) P(A and B) = P(A) * P(B)
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Example You are given the following information about 25 members of a sports club. Heavy Drinker Moderate Drinker Non- Drinker Women17210 Men48315 Total515525
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Example You are given the following information about 25 members of a sports club. Choose a person at random. a) What is the probability the chosen person is a women or non drinker? b) What is the probability the chosen person is a man and a moderate drinker? c) If the chosen person is a man, what is the probability he is a heavy drinker? Heavy Drinker Moderate Drinker Non- Drinker Women17210 Men48315 Total515525
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Example You are given the following information about 25 members of a sports club. Choose a person at random. d) Are the events women and heavy drinker, mutually exclusive? e) Find two independent events. Heavy Drinker Moderate Drinker Non- Drinker Women17210 Men48315 Total515525
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Example You are given the following information about 25 members of a sports club. Choose a person at random. a) What is the probability the chosen person is a women or non drinker?13/25=.52 b) What is the probability the chosen person is a man and a moderate drinker? c) If the chosen person is a man, what is the probability he is a heavy drinker? Heavy Drinker Moderate Drinker Non- Drinker Women17210 Men48315 Total515525
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Example You are given the following information about 25 members of a sports club. Choose a person at random. a) What is the probability the chosen person is a women or non drinker?13/25=.52 b) What is the probability the chosen person is a man and a moderate drinker? 8/25=.32 c) If the chosen person is a man, what is the probability he is a heavy drinker? Heavy Drinker Moderate Drinker Non- Drinker Women17210 Men48315 Total515525
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Example You are given the following information about 25 members of a sports club. Choose a person at random. a) What is the probability the chosen person is a women or non drinker?13/25=.52 b) What is the probability the chosen person is a man and a moderate drinker? 8/25=.32 c) If the chosen person is a man, what is the probability he is a heavy drinker?4/15=.267 Heavy Drinker Moderate Drinker Non- Drinker Women17210 Men48315 Total515525
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Example You are given the following information about 25 members of a sports club. Choose a person at random. d) Are the events women and heavy drinker, mutually exclusive? e) Find two independent events. Heavy Drinker Moderate Drinker Non- Drinker Women17210 Men48315 Total515525
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Example You are given the following information about 25 members of a sports club. Choose a person at random. d) Are the events women and heavy drinker, mutually exclusive? NO e) Find two independent events. Heavy Drinker Moderate Drinker Non- Drinker Women17210 Men48315 Total515525
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Example You are given the following information about 25 members of a sports club. Choose a person at random. d) Are the events women and heavy drinker, mutually exclusive? NO e) Find two independent events. Non-drinker and Women Non-drinker and Man Heavy Drinker Moderate Drinker Non- Drinker Women17210 Men48315 Total515525
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Tree diagrams One can keep track of conditional probabilities on a tree diagram. For example:
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Tree diagrams Three manufacture locations provide computer chips for Nortel Networks. Plants in Toronto, Vancouver and Montreal supply 60%, 25% and 15% respectively of computer chips for Nortel. The quality control department determines: 1% of chips from the plant in Toronto are defective 2% of chips from the plant in Van. are defective 2% of chips from the plant in Montreal are defective. - What is the probability that one of these chips is defective?
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Tree diagrams
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TO VAN MON
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Tree diagrams TO VAN MON 0.60 0.25 0.15
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Tree diagrams TO VAN MON 0.60 0.25 0.15 D N D – Defected Chip N – Not Defected
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Tree diagrams TO VAN MON 0.60 0.25 0.15 D N D – Defected Chip N – Not Defected D N D N
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Tree diagrams TO VAN MON 0.60 0.25 0.15 D N D – Defected Chip N – Not Defected D N D N 0.01 0.99 0.02 0.98 0.02 0.98
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Tree diagrams TO VAN MON 0.60 0.25 0.15 D N D – Defected Chip N – Not Defected D N D N 0.01 0.99 0.02 0.98 0.02 0.98 Now identify the events of interest and obtain probabilities of them by multiplying.
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Tree diagrams Three manufacture locations provide computer chips for Nortel Networks. Plants in Toronto, Vancouver and Montreal supply 60%, 25% and 15% respectively of computer chips for Nortel. The quality control department determines: 1% of chips from the plant in Toronto are defective 2% of chips from the plant in Van. are defective 2% of chips from the plant in Montreal are defective. - What is the probability that one of these chips is defective?
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Tree diagrams TO VAN MON 0.60 0.25 0.15 D N D – Defected Chip N – Not Defected D N D N 0.01 0.99 0.02 0.98 0.02 0.98 Now identify the events of interest and obtain probabilities of them by multiplying.
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Tree diagrams TO VAN MON 0.60 0.25 0.15 D N D – Defected Chip N – Not Defected D N D N 0.01 0.99 0.02 0.98 0.02 0.98 Now identify the events of interest and obtain probabilities of them by multiplying.
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Tree diagrams TO VAN MON 0.60 0.25 0.15 D N D – Defected Chip N – Not Defected D N D N 0.01 0.99 0.02 0.98 0.02 0.98 Now identify the events of interest and obtain probabilities of them by multiplying. 0.60*0.01=0.006 0.25*0.02=0.005 0.15*0.02=0.003
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Tree diagrams TO VAN MON 0.60 0.25 0.15 D N D – Defected Chip N – Not Defected D N D N 0.01 0.99 0.02 0.98 0.02 0.98 Now identify the events of interest and obtain probabilities of them by multiplying. The answer is the sum of these probabilities. 0.60*0.01=0.006 0.25*0.02=0.005 0.15*0.02=0.003
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Tree diagrams TO VAN MON 0.60 0.25 0.15 D N D – Defected Chip N – Not Defected D N D N 0.01 0.99 0.02 0.98 0.02 0.98 Answer: 0.006+0.005+0.003=0.014 OR 1.4% 0.60*0.01=0.006 0.25*0.02=0.005 0.15*0.02=0.003
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Tree diagrams Three manufacture locations provide computer chips for Nortel Networks. Plants in Toronto, Vancouver and Montreal supply 60%, 25% and 15% respectively of computer chips for Nortel. The quality control department determines: 1% of chips from the plant in Toronto are defective 2% of chips from the plant in Van. are defective 2% of chips from the plant in Montreal are defective. - What is the probability that a defective chip came from Montreal?
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Tree diagrams TO VAN MON 0.60 0.25 0.15 D N D – Defected Chip N – Not Defected D N D N 0.01 0.99 0.02 0.98 0.02 0.98 0.60*0.01=0.006 0.25*0.02=0.005 0.15*0.02=0.003
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Tree diagrams TO VAN MON 0.60 0.25 0.15 D N D – Defected Chip N – Not Defected D N D N 0.01 0.99 0.02 0.98 0.02 0.98 0.60*0.01=0.006 0.25*0.02=0.005 0.15*0.02=0.003
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Tree diagrams TO VAN MON 0.60 0.25 0.15 D N D – Defected Chip N – Not Defected D N D N 0.01 0.99 0.02 0.98 0.02 0.98 0.60*0.01=0.006 0.25*0.02=0.005 0.15*0.02=0.003
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Tree diagrams TO VAN MON 0.60 0.25 0.15 D N D – Defected Chip N – Not Defected D N D N 0.01 0.99 0.02 0.98 0.02 0.98 0.60*0.01=0.006 0.25*0.02=0.005 0.15*0.02=0.003
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Descriptive Phrases Descriptive Phrases require special care! – At most – At least – No more than – No less than
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Problems Problems 3.66, 3.68, 3.70, 3.75, 3.80, 3.87
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Review Conditional Probabilities Independent events Multiplication Rule Tree Diagrams
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34 Homework Review Chapter 3 Read Chapter 4.1-4.4
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