Review Conditional Probability Denoted PAB it represents the

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)

Example You are given the following information about 25 members of a sports club. Heavy Drinker Women Men Total 1 4 5 Moderate Drinker 7 8 15 Non. Drinker 2 3 5 10 15 25

Example You are given the following information about 25 members of a sports club. Choose a person at random. Heavy Drinker Women Men Total 1 4 5 Moderate Drinker 7 8 15 Non. Drinker 2 3 5 10 15 25 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?

Example You are given the following information about 25 members of a sports club. Choose a person at random. Heavy Drinker Women Men Total 1 4 5 Moderate Drinker 7 8 15 Non. Drinker 2 3 5 10 15 25 d) Are the events women and heavy drinker, mutually exclusive? e) Find two independent events.

Example You are given the following information about 25 members of a sports club. Choose a person at random. Heavy Drinker Women Men Total 1 4 5 Moderate Drinker 7 8 15 Non. Drinker 2 3 5 10 15 25 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?

Example You are given the following information about 25 members of a sports club. Choose a person at random. Heavy Drinker Women Men Total 1 4 5 Moderate Drinker 7 8 15 Non. Drinker 2 3 5 10 15 25 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?

Example You are given the following information about 25 members of a sports club. Choose a person at random. Heavy Drinker Women Men Total 1 4 5 Moderate Drinker 7 8 15 Non. Drinker 2 3 5 10 15 25 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

Example You are given the following information about 25 members of a sports club. Choose a person at random. Heavy Drinker Women Men Total 1 4 5 Moderate Drinker 7 8 15 Non. Drinker 2 3 5 10 15 25 d) Are the events women and heavy drinker, mutually exclusive? e) Find two independent events.

Example You are given the following information about 25 members of a sports club. Choose a person at random. Heavy Drinker Women Men Total 1 4 5 Moderate Drinker 7 8 15 Non. Drinker 2 3 5 10 15 25 d) Are the events women and heavy drinker, mutually exclusive? NO e) Find two independent events.

Example You are given the following information about 25 members of a sports club. Choose a person at random. Heavy Drinker Women Men Total 1 4 5 Moderate Drinker 7 8 15 Non. Drinker 2 3 5 10 15 25 d) Are the events women and heavy drinker, mutually exclusive? NO e) Find two independent events. Non-drinker and Women Non-drinker and Man

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

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?

Tree diagrams

Tree diagrams TO VAN MON

Tree diagrams TO 0. 60 0. 25 VAN 0. 15 MON

Tree diagrams D – Defected Chip D N – Not Defected TO 0. 60 0. 25 VAN 0. 15 MON N

Tree diagrams D – Defected Chip D N – Not Defected TO 0. 60 0. 25 N D VAN N D 0. 15 MON N

Tree diagrams D – Defected Chip N – Not Defected TO 0. 60 0. 25 0. 01 D 0. 99 N 0. 02 D VAN 0. 98 0. 02 0. 15 N D MON 0. 98 N

Tree diagrams D – Defected Chip N – Not Defected TO 0. 60 0. 25 0. 01 D 0. 99 N 0. 02 D VAN 0. 98 0. 02 0. 15 N D MON 0. 98 N Now identify the events of interest and obtain probabilities of them by multiplying.

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?

Tree diagrams D – Defected Chip N – Not Defected TO 0. 60 0. 25 0. 01 D 0. 99 N 0. 02 D VAN 0. 98 0. 02 0. 15 N D MON 0. 98 N Now identify the events of interest and obtain probabilities of them by multiplying.

Tree diagrams D – Defected Chip N – Not Defected TO 0. 60 0. 25 0. 01 D 0. 99 N 0. 02 D VAN 0. 98 0. 02 0. 15 N D MON 0. 98 N Now identify the events of interest and obtain probabilities of them by multiplying.

Tree diagrams D – Defected Chip N – Not Defected TO 0. 60 0. 25 0. 60*0. 01=0. 006 0. 01 D 0. 99 N 0. 02 D 0. 25*0. 02=0. 005 N D 0. 15*0. 02=0. 003 VAN 0. 98 0. 02 0. 15 MON 0. 98 N Now identify the events of interest and obtain probabilities of them by multiplying.

Tree diagrams D – Defected Chip N – Not Defected TO 0. 60 0. 25 0. 60*0. 01=0. 006 0. 01 D 0. 99 N 0. 02 D 0. 25*0. 02=0. 005 N D 0. 15*0. 02=0. 003 VAN 0. 98 0. 02 0. 15 MON 0. 98 N Now identify the events of interest and obtain probabilities of them by multiplying. The answer is the sum of these probabilities.

Tree diagrams D – Defected Chip N – Not Defected TO 0. 60 0. 25 0. 60*0. 01=0. 006 0. 01 D 0. 99 N 0. 02 D 0. 25*0. 02=0. 005 N D 0. 15*0. 02=0. 003 VAN 0. 98 0. 02 0. 15 MON 0. 98 N Answer: 0. 006+0. 005+0. 003=0. 014 1. 4% OR

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?

Tree diagrams D – Defected Chip N – Not Defected TO 0. 60 0. 25 0. 60*0. 01=0. 006 0. 01 D 0. 99 N 0. 02 D 0. 25*0. 02=0. 005 N D 0. 15*0. 02=0. 003 VAN 0. 98 0. 02 0. 15 MON 0. 98 N

Tree diagrams D – Defected Chip N – Not Defected TO 0. 60 0. 25 0. 60*0. 01=0. 006 0. 01 D 0. 99 N 0. 02 D 0. 25*0. 02=0. 005 N D 0. 15*0. 02=0. 003 VAN 0. 98 0. 02 0. 15 MON 0. 98 N

Tree diagrams D – Defected Chip N – Not Defected TO 0. 60 0. 25 0. 60*0. 01=0. 006 0. 01 D 0. 99 N 0. 02 D 0. 25*0. 02=0. 005 N D 0. 15*0. 02=0. 003 VAN 0. 98 0. 02 0. 15 MON 0. 98 N

Tree diagrams D – Defected Chip N – Not Defected TO 0. 60 0. 25 0. 60*0. 01=0. 006 0. 01 D 0. 99 N 0. 02 D 0. 25*0. 02=0. 005 N D 0. 15*0. 02=0. 003 VAN 0. 98 0. 02 0. 15 MON 0. 98 N

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

Problems 3. 66, 3. 68, 3. 70, 3. 75, 3. 80, 3. 87

Review • • Conditional Probabilities Independent events Multiplication Rule Tree Diagrams

Homework • Review Chapter 3 • Read Chapter 4. 1 -4. 4 34