1. 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)

2. Example You are given the following information about 25 members of a sports club. Heavy Drinker Moderate Drinker Non- Drinker Women17210 Men48315 Total515525

3. 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

4. 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

5. 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

6. 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

7. 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

8. 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

9. 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

10. 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

11. Tree diagrams One can keep track of conditional probabilities on a tree diagram. For example:

12. 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?

13. Tree diagrams

14. TO VAN MON

15. Tree diagrams TO VAN MON 0.60 0.25 0.15

16. Tree diagrams TO VAN MON 0.60 0.25 0.15 D N D – Defected Chip N – Not Defected

17. Tree diagrams TO VAN MON 0.60 0.25 0.15 D N D – Defected Chip N – Not Defected D N D N

18. 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

19. 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.

20. 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?

21. 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.

22. 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.

23. 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

24. 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

25. 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

26. 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?

27. 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

28. 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

29. 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

30. 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

31. Descriptive Phrases Descriptive Phrases require special care! – At most – At least – No more than – No less than

32. Problems Problems 3.66, 3.68, 3.70, 3.75, 3.80, 3.87

33. Review Conditional Probabilities Independent events Multiplication Rule Tree Diagrams

34. 34 Homework Review Chapter 3 Read Chapter 4.1-4.4