Statistics and Modelling Course 2011. Topic: Introduction to Probability Part of Achievement Standard 90643 Solve straightforward problems involving probability.

Statistics and Modelling Course 2011

Topic: Introduction to Probability Part of Achievement Standard 90643 Solve straightforward problems involving probability 4 Credits Externally Assessed NuLake Pages 102  130

Lesson 1: Sample space and combined events Sample space. Complementary events. The intersection between 2 events. The union between 2 events. Use new edition of Sigma (photocopy): Do Sigma Ex. 6.02 (p105) – Venn Diagrams. Ex. 6.03 (p108) Q2 & 3 – Contingency Tables

Basic Probability Sample space : set of all possible outcomes of an experiment. Event: any subset of sample space. E.g. A video game involves a player having 3 shots at a target. If he gets 2 or more hits, he gets an extra bonus shot, and if after that he has 3 or more hits, he gets a final shot. (a) Show the possible outcomes on a tree diagram. (b) List the sample space.

Population: Year 13 SAM class at STC DRAW ON BOARD ALONGSIDE PROJECTOR IMAGE EventTake Calculus (C) Don’t take Calc. (C`) TOTAL Take Physics (Ph) Don’t take physics (Ph`) TOTAL

Venn Diagrams – represent events within a Sample Space (s) Takes Calc. P(C) = Sample Space – the set of all possible outcomes. All possible outcomes – probabilities add to 1 (i.e. 100%). Does NOT take calc. P(C ’ ) = 1- p(C) =

Complementary Events (notes) For any event “A” A: Event A occurs. A’: The Complement of A. Event A does NOT occur. where P(A’) = 1 – P(A) Example: If we select a person at random from this class, the events: “This class member takes Calculus”, C and “This class member does NOT take Calculus”, C` are complements of each other. P( C’) = 1 – P(C) = 1 – =

THE INTERSECTION “A” AND “B” Takes Calc (C) Takes Physics (Ph)

Takes Physics (Ph) Takes Calc (C) THE INTERSECTION “A” AND “B”

Takes Calc AND Physics (C  Ph) P(C  Ph) =

THE UNION “A” OR “B” (or both) Takes Calc OR Physics (or both) (C  Ph) P(C  Ph) =

THE UNION “A” OR “B” (or both) P(C)

THE UNION “A” OR “B” (or both) P(Ph) P(C) +

THE UNION “A” OR “B” (or both) P(C  Ph) –

THE UNION “A” OR “B” (or both) Takes Calc OR Physics (or both) (C  Ph) So for any 2 events A and B: P(A  B) = P(A) + P(B) – P(A  B) E.g. to find the union of a student chosen at random from this class taking either Calc or Physics (or both), we add the individual probabilities of the events ‘takes Calc’ and ‘takes Physics’, then subtract the intersection (the overlap).

NOTES: For any 2 events A and B : The intersection – BOTH A and B happen P(A  B) = P(A) + P(B) – P(A  B) The union – A or B happen (or both): P(A  B) = P(A) + P(B) – P(A  B) E.g. to find the union of a student chosen at random from this class taking either Calc or Physics (or both), we add the individual probabilities of the events ‘takes Calc’ and ‘takes Physics’, then subtract the intersection (the overlap), so that we don’t count it twice. E.g. The probability that a randomly chosen member of this class takes both calculus and physics.

Venn diagram Do Sigma pg. 6 – Ex. 1.2 (old version) pg. 105 – Ex. 6.02 (new version) *HW: Handout (contingency tables)

Lesson 3: Mutually exclusive & independent events Mutually exclusive events. Independence. 1. Notes on mutual exclusivity & independence. 2. Do Sigma (old) – Ex. 1.3 (pg. 9) Or Sigma (new) – Ex. 6.04 (pg. 111) 3. HW: Probability assignment (due Thurs.)

Mutually Exclusive Events EventTake Geography (G) Don’t take Geography (G`) TOTAL Take Physics (Ph) Don’t take Physics (Ph`) TOTAL

Mutually Exclusive Events Takes Geography (G) Takes Physics (Ph) There is NO OVERLAP. If one happens, the other can’t! If someone takes Geo, then he can’t take Physics. If someone takes Physics, then he can’t take Geo. So P(Geo  Ph) = ?

Mutually Exclusive Events Takes Geography (G) Takes Physics (Ph) There is NO OVERLAP. If one happens, the other can’t! If someone takes Geo, then he can’t take Physics. If someone takes Physics, then he can’t take Geo. So P(Geo  Ph) = 0

Mutually Exclusive Events (notes) If two events are Mutually Exclusive, it means that if one happens, the other cannot. So they cannot both occur. At STC in Year 13, you cannot take both Geography and Physics. The two events “Takes 13Geo” and “Takes 13Physics” are Mutually Exclusive. Test whether 2 events are mutually exclusive by finding whether or not there is any intersection between them P(A  B). Can they both occur? 2 events A and B are mutually exclusive if and only if P(A  B) = 0

Independence

If two events are Independent, it means that the event that one has occurred does NOT alter the probability of the other occurring. Examples of two events that are independent: Examples of two events that are not independent:

How to test for independence:  Any 2 events A and B are independent if and only if: P(A  B) = P(A) × P(B)  If P(A  B) ≠ P(A) × P(B), then events A and B are NOT Independent. Work – finish for HW: 1. Do Sigma pg. 9 – Ex. 1.3 (old version) pg. 111 – Ex. 6.04 (new version).

More combined events examples Do worksheet of past NCEA qs (involves unions, intersections, mutually exclusive events and the ‘neither’ event – using Venn Diagrams and Contingency Tables).

Lesson 4: Tree diagrams 1 Use tree diagrams to calculate probabilities. STARTER: The “Neither” event. Do NuLake: Tree Diagrams –Pg. 115  120 - Q16  26(h) Complete for HW.

The ‘Neither’ event

2006 NCEA exam question Rewa asked 150 randomly chosen students what programmes they watched last night. 90 watched Shortland Street, 50 NZ Idol, and 30 had watched both. What is the probability that a randomly chosen student had watched neither Shortland Street nor NZ Idol last night?

Next: Tree Diagrams – Do NuLake pg. 115  120

P(A υ B) The ‘NEITHER’ event P(A` ∩ B`)

P(A υ B) The ‘NEITHER’ event “Nor” = 1- “Or” P(Neither A nor B) = 1 – P(A υ B) This can be easily observed on a contingency table (see previous example). P(Neither A nor B) is written as P(A` ∩ B`), the probability that event A doesn’t occur AND event B doesn’t occur. P(A` ∩ B`)

Lesson 5: Tree diagrams 2 Practice and develop more confidence with using Tree Diagrams to solve problems involving probability. Do Sigma pg. 12 – Ex. 1.4 (old version): Q1-14 or, in new edition, pg. 116 – Ex.6.05 * Extension people: Do new edition – pg. 116 – Ex. 6.05: Do Q3  14, 15(Exc), then read infinite probabilities example at bottom of p115, then do Q16-20 (all Exc).

Lesson 6: Conditional probability 1. Intro to Conditional Probability Introduction to Conditional Probability and its notation and formula. Link to Tree Diagrams. Do NuLake pg. 123  126 Or Sigma Ex. 6.01, 6.02

Conditional Probability Conditional probability means the probability of a particular event occurring GIVEN that some other event has occurred. E.g. the probability that a randomly chosen person in this class has a car, GIVEN that we know he has a job. Once we know that a particular event has occurred our sample space of possibilities is reduced.

Conditional Probability NUMBER IN CLASS TODAY = ____ Q. Stand up if you have a job. n(Job) = P(Job) = Q. Of those standing, RAISE YOUR HAND if you ALSO own a car. n(Job  Car) = P(Job  Car) =

Conditional Probability NUMBER IN CLASS TODAY = ____ Q. Stand up if you have a job. n(Job) = P(Job) = Q. Of those standing, RAISE YOUR HAND if you ALSO own a car. n(Job  Car) = P(Job  Car) = The probability that a randomly selected member of this class has a car GIVEN that he has a job is given by: P( Car ᅵ Job ) = Number standing up with hands up Total number standing

Conditional Probability NUMBER IN CLASS TODAY = ____ Q. Stand up if you have a job. n(Job) = P(Job) = Q. Of those standing, RAISE YOUR HAND if you ALSO own a car. n(Job  Car) = P(Job  Car) = The probability that a randomly selected member of this class has a car GIVEN that he has a job is given by: P( Car ᅵ Job ) = Number standing up with hands up = n( ? ) Total number standing n( ? )

Conditional Probability NUMBER IN CLASS TODAY = ____ Q. Stand up if you have a job. n(Job) = P(Job) = Q. Of those standing, RAISE YOUR HAND if you ALSO own a car. n(Job  Car) = P(Job  Car) = The probability that a randomly selected member of this class has a car GIVEN that he has a job is given by: P( Car ᅵ Job ) = Number standing up with hands up = n(Job  Car) Total number standing n(Job) Show this information on a contingency table. HAVE A GO: Do NuLake pg. 123 - JUST Question 29.

Conditional Probability on a Venn diagram P(A  B`) P(B  A`) P(A  B) P(A’  B’)

P(A  B`) P(A  B) P(A’) Conditional Probability on a Venn diagram

We KNOW Event A has occurred Conditional Probability on a Venn diagram

P( BᅵA) We KNOW Event A has occurred Conditional Probability on a Venn diagram

P( BᅵA) We KNOW Event A has occurred RULE: The conditional probability that event B occurs if we already know that A has happened is written as P(B ᅵ A). We say “the probability of B, given A”. P(B ᅵ A)= P(A  B) P(A) Something to ponder: If A and B are independent, then P(B ᅵ A) = P(___) Conditional Probability on a Venn diagram HAVE A GO: Do NuLake pg. 124  126 You MUST do: Q30-40. * Extension: Q41.

Lesson 7: Conditional probability 2. Watch youtube clip on conditional probability: http://www.youtube.com/watch?v=4PwnvqGEHoU Link with tree-diagrams. Look at one together. Which probabilities on the diagram are conditional probabilities? Practice solving problems involving conditional probability using Venn diagrams. 1.Go over HW qs from NuLake & finish these off. 2. Faster people move on to NuLake pages 130  134.

Lesson 8 (if time): Conditional probability 3. Problem-solving: The Base Rate Fellacy. Last lesson to practice conditional probability problems and sort out any difficulties. 1.) Finish NuLake pages 130  134 2.) Sigma (new): Pg. 176 – Ex. 9.04 + answer qs for Unlawful Activity investigation. Extension: 2010 NCEA exam (AS90643) – final question (hard Venn diagram problem).

The base-rate fellacy

In a certain town there are two types of taxis; 90% are Blue Cabs and 10% are Red Cabs.

The base-rate fellacy In a certain town there are two types of taxis; 90% are Blue Cabs and 10% are Red Cabs. A taxi is involved in a hit-and- run accident.

The base-rate fellacy In a certain town there are two types of taxis; 90% are Blue Cabs and 10% are Red Cabs. A taxi is involved in a hit-and- run accident. One witness says it was a Red Cab. The court tests this witness’s ability to distinguish between Blue and Red cabs under identical conditions and concludes that the witness is right 80% of the time. Question: What is the probability that the taxi was a Red Cab? Lawyers, judges, doctors and even maths teachers have claimed that the answer is 80%. Yet this is incorrect.

The base-rate fellacy In a certain town there are two types of taxis; 90% are Blue Cabs and 10% are Red Cabs. A taxi is involved in a hit-and- run accident. One witness says it was a Red Cab. The court tests this witness’s ability to distinguish between Blue and Red cabs under identical conditions and concludes that the witness is right 80% of the time. Question: What is the probability that the taxi was a Red Cab? Lawyers, judges, doctors and even maths teachers have claimed that the answer is 80%. Yet this is incorrect. Can you calculate the true probability? And can you explain why 80% is wrong? HINT: Draw a probability tree.SPOILER WARNING!!!!!!! OK. The answer is… 0.3077(4sf) So it’s actually more likely that the witness got it wrong and the taxi was blue! Use a probability tree or otherwise to show that this is the correct answer. Do Sigma (old version) pg. 19 – Ex. 1.6 Or new version pg. 176 – Ex. 9.04 (M&E)

Extra for experts lesson – Bayes’ Theorem. Practice solving problems involving conditional probability.

Exploring Conditional Probability – swine flu At one stage last year, Ministry of Health figures indicate that 2% of Christchurch high school students had Swine Flu. A particular medical test could correctly identify whether a patient had Swine Flu 90% of the time.

Exploring Conditional Probability – swine flu At one stage last year, Ministry of Health figures indicate that 2% of Christchurch high school students had Swine Flu. At the time, health clinics conducted a test that accurately detected the virus for 90% of people with it. However the test also indicated the disease for 12.5% of people without it (the false positives). Draw a tree diagram showing 2 variables: 1.whether a randomly chosen high school student actually has swine flu (F) or not (F`), and for each of these scenarios… 2.whether the test says that they have it (T) or not (T`). Q1: If a person actually has Swine Flu, what is P(T|F), the probability that the test will say that they have it? Q2: What is the probability P(T) that any randomly selected student tests positive for Swine Flu.

At one stage last year, Ministry of Health figures indicate that 2% of Christchurch high school students had Swine Flu. A particular medical test could correctly identify whether a patient had Swine Flu 90% of the time. This means that 90% of people with Swine Flu would test positive for it, while 90% of people without Swine Flu would test negative for it. Draw a tree diagram showing 2 variables: 1.whether a randomly chosen high school student actually has swine flu (F) or not (F`), and for each of these scenarios… 2.whether the test says that they have it (T) or not (T`). Q1: If a person actually has Swine Flu, what is P(T|F), the probability that the test will say that they have it? Q2: What is the probability P(T) that any randomly selected student tests positive for Swine Flu. Q3: If a person has tested positive for Swine Flu, what is P(F|T), the probability that he/she actually has Swine Flu? Q4: Write a formula for P(F|T) in terms of P(F), P(T) and P(T|F).

Q1: If a person actually has Swine Flu, what is P(T|F), the probability that the test will say that they have it? Q2: What is the probability P(T) that any randomly selected student tests positive for Swine Flu. Q3: If a person has tested positive for Swine Flu, what is P(F|T), the probability that he/she actually has Swine Flu? Q4: Write a formula for P(F|T) in terms of P(F), P(T) and P(T|F). Q5: Generalise your formula so that it applies to any 2 events, A and B where event A happens before B. Your formula is for calculating the probability that event A happened, given that B has happened, based on P(A), P(B) and P(B |A ). PROBLEMS: Do Statistics & Modelling NCEA Level 3 Revision Guide: Probability: Q1(v), Q25, Q26(answer in back wrong), Schol qs from 2008 & 2009.

Q1: If a person actually has Swine Flu, what is P(T|F), the probability that the test will say that they have it? Q2: What is the probability P(T) that any randomly selected student tests positive for Swine Flu. Q3: If a person has tested positive for Swine Flu, what is P(F|T), the probability that he/she actually has Swine Flu? Q4: Write a formula for P(F|T) in terms of P(F), P(T) and P(T|F). Q5: Generalise your formula so that it applies to any 2 events, A and B where event A happens before B. Your formula is for calculating the probability that event A happened, given that B has happened, based on P(A), P(B) and P(B |A ). PROBLEMS: Do Statistics & Modelling NCEA Level 3 Revision Guide: Probability: Q1(v), Q25, Q26(answer in back wrong), Schol qs from 2008 & 2009. BAYES’ THEOREM: For 2 events A and B, If P(A), P(B) and P(B ᅵ A) are known, then: P(A ᅵ B)= P(B ᅵ A).P(A) P(B)

Q1: If a person actually has Swine Flu, what is P(T|F), the probability that the test will say that they have it? Q2: What is the probability P(T) that any randomly selected student tests positive for Swine Flu. Q3: If a person has tested positive for Swine Flu, what is P(F|T), the probability that he/she actually has Swine Flu? Q4: Write a formula for P(F|T) in terms of P(F), P(T) and P(T|F). Q5: Generalise your formula so that it applies to any 2 events, A and B where event A happens before B. Your formula is for calculating the probability that event A happened, given that B has happened, based on P(A), P(B) and P(B |A ). PROBLEMS: Do Statistics & Modelling NCEA Level 3 Revision Guide: Probability: Q1(v), Q25, Q26(answer in back wrong), Schol qs from 2008 & 2009. BAYES’ THEOREM: For 2 events A and B, If P(A), P(B) and P(B ᅵ A) are known, then: Proof: P(A ᅵ B)= P(B ᅵ A).P(A) P(B)

Q1: If a person actually has Swine Flu, what is P(T|F), the probability that the test will say that they have it? Q2: What is the probability P(T) that any randomly selected student tests positive for Swine Flu. Q3: If a person has tested positive for Swine Flu, what is P(F|T), the probability that he/she actually has Swine Flu? Q4: Write a formula for P(F|T) in terms of P(F), P(T) and P(T|F). Q5: Generalise your formula so that it applies to any 2 events, A and B where event A happens before B. Your formula is for calculating the probability that event A happened, given that B has happened, based on P(A), P(B) and P(B |A ). PROBLEMS: Do Statistics & Modelling NCEA Level 3 Revision Guide: Probability: Q1(v), Q25, Q26(answer in back wrong), Schol qs from 2008 & 2009. BAYES’ THEOREM: For 2 events A and B, If P(A), P(B) and P(B ᅵ A) are known, then: Proof: Since P(B ᅵ A) = P(A  B) P(A) P(A ᅵ B)= P(B ᅵ A).P(A) P(B)

Q1: If a person actually has Swine Flu, what is P(T|F), the probability that the test will say that they have it? Q2: What is the probability P(T) that any randomly selected student tests positive for Swine Flu. Q3: If a person has tested positive for Swine Flu, what is P(F|T), the probability that he/she actually has Swine Flu? Q4: Write a formula for P(F|T) in terms of P(F), P(T) and P(T|F). Q5: Generalise your formula so that it applies to any 2 events, A and B where event A happens before B. Your formula is for calculating the probability that event A happened, given that B has happened, based on P(A), P(B) and P(B |A ). PROBLEMS: Do Statistics & Modelling NCEA Level 3 Revision Guide: Probability: Q1(v), Q25, Q26(answer in back wrong), Schol qs from 2008 & 2009. BAYES’ THEOREM: For 2 events A and B, If P(A), P(B) and P(B ᅵ A) are known, then: Proof: Since P(B ᅵ A) = P(A  B) P(A) So P(A  B) = P(B ᅵ A).P(A). P(A ᅵ B)= P(B ᅵ A).P(A) P(B)

Q1: If a person actually has Swine Flu, what is P(T|F), the probability that the test will say that they have it? Q2: What is the probability P(T) that any randomly selected student tests positive for Swine Flu. Q3: If a person has tested positive for Swine Flu, what is P(F|T), the probability that he/she actually has Swine Flu? Q4: Write a formula for P(F|T) in terms of P(F), P(T) and P(T|F). Q5: Generalise your formula so that it applies to any 2 events, A and B where event A happens before B. Your formula is for calculating the probability that event A happened, given that B has happened, based on P(A), P(B) and P(B |A ). PROBLEMS: Do Statistics & Modelling NCEA Level 3 Revision Guide: Probability: Q1(v), Q25, Q26(answer in back wrong), Schol qs from 2008 & 2009. BAYES’ THEOREM: For 2 events A and B, If P(A), P(B) and P(B ᅵ A) are known, then: Proof: Since P(B ᅵ A) = P(A  B) P(A) So P(A  B) = P(B ᅵ A).P(A). We can then substitute this into the conditional probability formula: P(A ᅵ B)= P(B ᅵ A).P(A) P(B)

Q1: If a person actually has Swine Flu, what is P(T|F), the probability that the test will say that they have it? Q2: What is the probability P(T) that any randomly selected student tests positive for Swine Flu. Q3: If a person has tested positive for Swine Flu, what is P(F|T), the probability that he/she actually has Swine Flu? Q4: Write a formula for P(F|T) in terms of P(F), P(T) and P(T|F). Q5: Generalise your formula so that it applies to any 2 events, A and B where event A happens before B. Your formula is for calculating the probability that event A happened, given that B has happened, based on P(A), P(B) and P(B |A ). PROBLEMS: Do Statistics & Modelling NCEA Level 3 Revision Guide: Probability: Q1(v), Q25, Q26(answer in back wrong), Schol qs from 2008 & 2009. BAYES’ THEOREM: For 2 events A and B, If P(A), P(B) and P(B ᅵ A) are known, then: Proof: Since P(B ᅵ A) = P(A  B) P(A) So P(A  B) = P(B ᅵ A).P(A). We can then substitute this into the conditional probability formula: P(A ᅵ B) = P(A  B) P(B) P(A ᅵ B)= P(B ᅵ A).P(A) P(B)

Q1: If a person actually has Swine Flu, what is P(T|F), the probability that the test will say that they have it? Q2: What is the probability P(T) that any randomly selected student tests positive for Swine Flu. Q3: If a person has tested positive for Swine Flu, what is P(F|T), the probability that he/she actually has Swine Flu? Q4: Write a formula for P(F|T) in terms of P(F), P(T) and P(T|F). Q5: Generalise your formula so that it applies to any 2 events, A and B where event A happens before B. Your formula is for calculating the probability that event A happened, given that B has happened, based on P(A), P(B) and P(B |A ). PROBLEMS: Do Statistics & Modelling NCEA Level 3 Revision Guide: Probability: Q1(v), Q25, Q26(answer in back wrong), Schol qs from 2008 & 2009. BAYES’ THEOREM: For 2 events A and B, If P(A), P(B) and P(B ᅵ A) are known, then: Proof: Since P(B ᅵ A) = P(A  B) P(A) So P(A  B) = P(B ᅵ A).P(A). We can then substitute this into the conditional probability formula: P(A ᅵ B) = P(A  B) = P(B ᅵ A).P(A) P(B) P(B) P(A ᅵ B)= P(B ᅵ A).P(A) P(B)

2006 Scholarship Question All incoming strawberries for making yoghurt are checked for defects. An automatic scanner is used, followed by a visual inspection. The scanner removes 90% of defective strawberries and the visual inspection process removes 70% of any remaining defective strawberries. From previous records, the company knows that 4% of incoming strawberries are defective. (a) Calculate the percentage of defective strawberries that are removed by the visual inspection process. (b) Calculate the probability that a strawberry that was found to be defective had been removed by the scanner. (c) If the scanner is correct 90% of the time, what is the probability that a strawberry registered as ‘defective’ by the scanner is actually defective.

2006 Scholarship Question All incoming strawberries for making yoghurt are checked for defects. An automatic scanner is used, followed by a visual inspection. The scanner removes 90% of defective strawberries and the visual inspection process removes 70% of any remaining defective strawberries. From previous records, the company knows that 4% of incoming strawberries are defective. (a) Calculate the percentage of defective strawberries that are removed by the visual inspection process.

ANSWER to (a) (a)7% of strawberries are removed by visual inspection.

2006 Scholarship Question All incoming strawberries for making yoghurt are checked for defects. An automatic scanner is used, followed by a visual inspection. The scanner removes 90% of defective strawberries and the visual inspection process removes 70% of any remaining defective strawberries. From previous records, the company knows that 4% of incoming strawberries are defective. (b) Calculate the probability that a strawberry that was found to be defective had been removed by the scanner.

ANSWER to (b) (a)P(Removed by scanner ᅵ Found defective)= 0.928 (3 sf)

Quiz – how much do you remember? Q1) Take two events “A” and “B”. Event A occurs 65% of the time, and Event “B” 32%. 75% of the time at least one of “A” or “B” occurs. (a)Draw a Contingency Table and a Venn Diagram showing this information. (b)Are events A and B mutually exclusive? Use statistical reasoning to justify your answer. (c)What percentage of the time do both A and B occur? (d)What percentage of the time do neither A nor B occur? (e)Are events A and B statistically independent? Use statistical reasoning to justify your answer. Read Sigma (new) pg. 110.

Quiz – how much do you remember? Q1) Take two events “A” and “B”. Event A occurs 65% of the time, and Event B 32%. 75% of the time at least one of “A” or “B” occurs. (a)Draw a Contingency Table and a Venn Diagram showing this information. (b)Are events A and B mutually exclusive? Use statistical reasoning to justify your answer. (c)What percentage of the time do both A and B occur? (d)What percentage of the time do neither A nor B occur? (e)Are events A and B statistically independent? Use statistical reasoning to justify your answer. Read Sigma (new) pg. 110.

Quiz – how much do you remember? Q1) Take two events “A” and “B”. Event A occurs 65% of the time, and Event B 32%. 75% of the time at least one of “A” or “B” occurs. (a)Draw a Contingency Table showing this information. EventEvent A occurs (A) Event A does not occur (A`) TOTAL Event B occurs (B) Event B does not occur (B`) TOTAL

EVENTS “A” and “B” A P(A) = 0.65 B P(B) = 0.32

Venn Diagram

A OR B (or both) (A υ B) P(A υ B) = 0.75

Quiz – how much do you remember? Q1) Take two events “A” and “B”. Event A occurs 65% of the time, and Event B 32%. 75% of the time, at least one of “A” or “B” occurs. (a)Draw a Contingency Table showing this information. (b)Are events A and B mutually exclusive? Use statistical reasoning to justify your answer. (c)What percentage of the time do both A and B occur? (d)What percentage of the time does neither A nor B occur? (e)Are events A and B statistically independent? Use statistical reasoning to justify your answer.

A P(A) = 0.65 B P(B) = 0.32 If “A” and “B” were mutually exclusive, then: P(A  B) = ? +?

B P(B) = 0.32 A P(A) = 0.65 If “A” and “B” were mutually exclusive, then: P(A  B) = P(A) + P(B)

B P(B) = 0.32 A P(A) = 0.65 If “A” and “B” were mutually exclusive, then: P(A  B) = P(A) + P(B) Why?

B P(B) = 0.32 A P(A) = 0.65 If “A” and “B” were mutually exclusive, then: P(A  B) = P(A) + P(B) Why? Because there would be NO INTERSECTION

B P(B) = 0.32 A P(A) = 0.65 If “A” and “B” were mutually exclusive, then: P(A  B) = P(A) + P(B) Why? Because there would be NO INTERSECTION If 2 events A & B are mutually exclusive then P(A  B) = 0.

B P(B) = 0.32 A P(A) = 0.65 If “A” and “B” were mutually exclusive, then: P(A  B) = P(A) + P(B) - P(A  B) Why? Because there would be NO INTERSECTION If 2 events A & B are mutually exclusive then P(A  B) = 0.

B P(B) = 0.32 A P(A) = 0.65 If “A” and “B” were mutually exclusive, then: P(A  B) = P(A) + P(B) - 0 Why? Because there would be NO INTERSECTION If 2 events A & B are mutually exclusive then P(A  B) = 0.

B P(B) = 0.32 A P(A) = 0.65 If “A” and “B” were mutually exclusive, then: P(A  B) = P(A) + P(B) - 0 Why? Because there would be NO INTERSECTION If 2 events A & B are mutually exclusive then P(A  B) = 0. Is this true in this e.g.?

A P(A) = 0.65 B P(B) = 0.32 If “A” and “B” were mutually exclusive, then: P(A  B) = P(A) + P(B) - 0 Why? Because there would be NO INTERSECTION If 2 events A & B are mutually exclusive then P(A  B) = 0. Is this true in this e.g.? P(A  B) = 0.75 P(A  B) = P(A) + P(B) – P(A  B)

A P(A) = 0.65 B P(B) = 0.32 If “A” and “B” were mutually exclusive, then: P(A  B) = P(A) + P(B) Why? Because there would be NO INTERSECTION If 2 events A & B are mutually exclusive then P(A  B) = 0. Is this true in this e.g.? P(A  B) = 0.75 P(A  B) = P(A) + P(B) – P(A  B)

P(A υ B) = 0.75 (d) What percentage of the time does neither A nor B occur?

P(A υ B) = 0.75 (d) What percentage of the time does neither A nor B occur? “Nor” = 1- “Or” P(Neither A nor B) = 1 – P(A υ B) = 1 – 0.75 = 0.25

Q2) Some year 13 students on a school trip are travelling past some road works and one leans out the window and tries to grab the road cones as they pass. There are 3 cones in a row. For each cone, the probability that he collects it is 0.2, that he just knocks it over is 0.55, and that he misses it altogether is 0.25. If he collects one he will stop, content with his prize. Otherwise he will keep trying. (a) Draw a probability tree for this situation.

Q2) Some year 13 students on a school trip are travelling past some road works and one leans out the window and tries to grab the road cones as they pass. There are 3 cones in a row. For each cone, the probability that he collects it is 0.2, that he just knocks it over is 0.55, and that he misses it altogether is 0.25. If he collects one he will stop, content with his prize. Otherwise he will keep trying. (a) Draw a probability tree for this situation. Calculate the probability that… (b) he knocks over all 3 cones. (c) he misses the first two cones, but collects the third.

Q2. Some year 13 students on a school trip are travelling past some road works and one leans out the window and tries to grab the road cones as they pass. There are 3 cones in a row. For each cone, the probability that he collects it is 0.2, that he just knocks it over is 0.55, and that he misses it altogether is 0.25. If he collects one he will stop, content with his prize. Otherwise he will keep trying. (c) he misses the first two cones, but collects the third. (d) he knocks over exactly two cones. (e) he collects one of the cones (call this event A)

Some year 13 students on a school trip are travelling past some road works and one leans out the window and tries to grab the road cones as they pass. There are 3 cones in a row. For each cone, the probability that he collects it is 0.2, that he just knocks it over is 0.55, and that he misses it altogether is 0.25. If he collects one he will stop, content with his prize. Otherwise he will keep trying. (e) he collects one of the cones (call this event A) (f) Give three examples of pairs of events that are mutually exclusive in this situation (think in terms of results – number of cones knocked over etc). (g) Another equally bright student tries to do the same thing on the other side of the van. He also has 3 cones in a row. For each cone, the probability that he collects it is 0.2063. He will also stop if he collects one. If the probability of both students collecting one of the cones is approximately 0.2440, are events A (student A collects one) and B (student B collects one) statistically independent? Use statistical reasoning to justify your answer.

Some year 13 students on a school trip are travelling past some road works and one leans out the window and tries to grab the road cones as they pass. There are 3 cones in a row. For each cone, the probability that he collects it is 0.2, that he just knocks it over is 0.55, and that he misses it altogether is 0.25. If he collects one he will stop, content with his prize. Otherwise he will keep trying. (e) he collects one of the cones (call this event A) (f) Describe two events that are mutually exclusive in this situation (think in terms of results – number of cones knocked over etc). (g) Another equally bright student tries to do the same thing on the other side of the van. He also has 3 cones in a row. For each cone, the probability that he collects it is 0.2063. He will also stop if he collects one. If the probability of both students collecting one of the cones is approx. 0.2440, are events A (student A collects one) and B (student B collects one) statistically independent? Use statistical reasoning to justify your answer.

Statistics and Modelling Course 2011. Topic: Introduction to Probability Part of Achievement Standard 90643 Solve straightforward problems involving probability.

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Statistics and Modelling Course 2011. Topic: Introduction to Probability Part of Achievement Standard 90643 Solve straightforward problems involving probability.

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Presentation on theme: "Statistics and Modelling Course 2011. Topic: Introduction to Probability Part of Achievement Standard 90643 Solve straightforward problems involving probability."— Presentation transcript:

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