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COUNTING & PROBABILITY

COUNTING & PROBABILITY

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Page 1: COUNTING & PROBABILITY

COUNTING & PROBABILITY

Page 2: COUNTING & PROBABILITY

TERMINOLOGY RECAP

o Probability lies between 0 (impossible event) and 1 (certain event)

o If all events are equally likely, then the probability of the event is:

๐‘ƒ ๐ธ๐‘ฃ๐‘’๐‘›๐‘ก =๐‘ƒ(๐‘†๐‘ข๐‘๐‘๐‘’๐‘ ๐‘ ๐‘“๐‘ข๐‘™ ๐ธ๐‘ฃ๐‘’๐‘›๐‘ก)

๐‘‡๐‘œ๐‘ก๐‘Ž๐‘™ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘’๐‘ฃ๐‘’๐‘›๐‘ก๐‘ 

oSample space is made up of all the possible events

Page 3: COUNTING & PROBABILITY

TERMINOLOGY RECAP

oSet notation is often used, where โˆช is the union of sets

-> operation is ORโˆฉ is the intersection of sets

-> operation is AND

Intersection and Union of Sets

Practicing Basic Set Notation

Page 4: COUNTING & PROBABILITY

TOOLS USED IN PROBABILITY

Venn DiagramsShow all possible logical relations between a finite collection of sets

Practicing Venn Diagrams with Actual

Numbers

Practicing Venn Diagrams with Set

Notation

Page 5: COUNTING & PROBABILITY

TOOLS USED IN PROBABILITY

Tree DiagramsHave branches which show thepossible outcomesof multiple types of events

Tree Diagrams

Page 6: COUNTING & PROBABILITY

TOOLS USED IN PROBABILITY

Contingency TableIs a two-way table consisting of 2 variables to summarize data and compare the variables.

+ 1 2 3 4 5 6

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12

Contingency Table for Die

Using Contingency

Tables in Probabilities with

Monopoly

Page 7: COUNTING & PROBABILITY

INDEPENDENT EVENTS

oThe outcome of the 1st event has NO effect on the outcome of the 2nd event

oProduct Rule for Independent Events:๐‘ƒ ๐ด ๐‘Ž๐‘›๐‘‘ ๐ต = ๐‘ƒ ๐ด ร— ๐‘ƒ ๐ต๐‘ƒ ๐ด โˆฉ ๐ต = ๐‘ƒ ๐ด ร— ๐‘ƒ ๐ต

oE.g. What is the probability of flipping a coin twice and getting Heads twice?

๐‘ƒ ๐ป ๐‘Ž๐‘›๐‘‘ ๐ป = ๐‘ƒ ๐ป ร— ๐‘ƒ ๐ป

=1

2ร—

1

2

=1

4

The Coin Toss

Page 8: COUNTING & PROBABILITY

Twelve numbers are dropped into a hat:

Event C = {๐‘“๐‘Ž๐‘๐‘ก๐‘œ๐‘Ÿ๐‘  ๐‘œ๐‘“ 6}

Event D = {๐‘“๐‘Ž๐‘๐‘ก๐‘œ๐‘Ÿ๐‘  ๐‘œ๐‘“ 9}

๐‘ƒ ๐ถ โˆฉ ๐ท = ๐‘ƒ ๐ถ ร— ๐‘ƒ ๐ท

Independent Events Venn Diagram

Basic Independent Test Example

Page 9: COUNTING & PROBABILITY

Use a tree diagram to determine the probability of getting two Heads and one Tail when tossing a fair coin three times?

Answer:3

8

Independent Events Tree Diagram

Independent Tree

Diagram Example

Page 10: COUNTING & PROBABILITY

Independent Events Question:

What is the probability of a couple having 3 male children?

๐‘ƒ ๐ต ๐‘Ž๐‘›๐‘‘ ๐ต ๐‘Ž๐‘›๐‘‘ ๐ต

= ๐‘ƒ ๐ต ร— ๐‘ƒ ๐ต ร— ๐‘ƒ(๐ต)

=1

2ร—1

2ร—1

2

=1

8

Page 11: COUNTING & PROBABILITY

Independent Events Question:

If the probability of having an accident is 0,25 and the probability of using a taxi is 0,64, then determine the probability of having an accident while using a taxi.

๐‘ƒ ๐‘Ž๐‘๐‘๐‘–๐‘‘๐‘’๐‘›๐‘ก ๐‘Ž๐‘›๐‘‘ ๐‘ข๐‘ ๐‘–๐‘›๐‘” ๐‘ก๐‘Ž๐‘ฅ๐‘–

= ๐‘ƒ ๐‘Ž๐‘๐‘๐‘–๐‘‘๐‘’๐‘›๐‘ก ร— ๐‘ƒ ๐‘ข๐‘ ๐‘–๐‘›๐‘” ๐‘Ž ๐‘ก๐‘Ž๐‘ฅ๐‘–

= 0,25 ร— 0,64

= 0,16Independent Probability

Questions

Page 12: COUNTING & PROBABILITY

DEPENDENT EVENTS

oThe outcome of the 1st event effects the outcome of the 2nd event

oCan use the Product Rule:๐‘ƒ ๐ด ๐‘Ž๐‘›๐‘‘ ๐ต = ๐‘ƒ ๐ด ร— ๐‘ƒ ๐ต๐‘ƒ ๐ด โˆฉ ๐ต = ๐‘ƒ ๐ด ร— ๐‘ƒ ๐ต

oE.g. What is the probability of drawing two Aces from a standard deck of cards, if the first card drawn is an Ace and is not replaced?

๐‘ƒ ๐ด๐‘๐‘’ ร— ๐‘ƒ ๐ด๐‘๐‘’ =4

52ร—

3

๐Ÿ“๐Ÿ=

1

221 A Pack of Cards

Page 13: COUNTING & PROBABILITY

Sam and Alex work together. Sam has 0,6 chance of being picked to relocate. If Alex gets picked to relocate, the chance that he says yes is 0,3. Use a tree diagram to determine the probability of the boss picking Sam and Sam saying โ€œyesโ€ to the relocation.

Dependent Events Tree Diagram

Dependent Tree Diagram

Example

Page 14: COUNTING & PROBABILITY

Dependent Events Question:

Use a tree diagram to determine the probability of choosing two blue sweets (without replacement), if there are 12 Green sweets and 9 Blue sweets.

Page 15: COUNTING & PROBABILITY

Dependent Events Question:

What is the probability of picking two blue marbles from a hat containing 4 blue; 8 red and 6 green marbles, if the first marble isnโ€™t replaced?

๐‘ƒ(๐ต ๐‘Ž๐‘›๐‘‘ ๐ต) = ๐‘ƒ(๐ต) + ๐‘ƒ(๐ต)

=4

18+

3

17

=61

153Dependent Events

QuestionsIdentifying Dependent vs Independent Events

Page 16: COUNTING & PROBABILITY

MUTUALLY EXCLUSIVE EVENTS

oEvents that cannot happen at the same time are mutually exclusive events

oSum Rule for Mutually Exclusive Events:๐‘ƒ ๐ด ๐‘œ๐‘Ÿ ๐ต = ๐‘ƒ ๐ด + ๐‘ƒ ๐ต๐‘ƒ(๐ด โˆช ๐ต) = ๐‘ƒ ๐ด + ๐‘ƒ ๐ต

oE.g. What is the probability of getting a 1 or a 5 when rolling a dice?

๐‘ƒ 1 ๐‘œ๐‘Ÿ 5 = ๐‘ƒ 1 + ๐‘ƒ 5

=1

6+

1

6

=2

6=

1

2

Page 17: COUNTING & PROBABILITY

MUTUALLY EXCLUSIVE EVENTS

oNote! ๐‘ƒ ๐ด ๐‘Ž๐‘›๐‘‘ ๐ต = 0๐‘ƒ ๐ด โˆฉ ๐ต = 0

oE.g. What is the probability of getting a 1 and a 5 when rolling a dice?

๐‘ƒ 1 ๐‘Ž๐‘›๐‘‘ 5 = 0

Practical Examples of Mutually Exclusive Events

Page 18: COUNTING & PROBABILITY

Mutually Exclusive Venn Diagram

Twelve numbers are dropped into a hat:

Event A = {๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ๐‘  โ‰ค 6}

Event B = {๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ๐‘  > 6}

๐‘Ž) ๐‘ƒ(๐ด โˆช ๐ต)

= ๐‘ƒ ๐ด + ๐‘ƒ ๐ต

๐‘) ๐‘ƒ(๐ด โˆฉ ๐ต)

= 0

Page 19: COUNTING & PROBABILITY

Mutually Exclusive Events Question:

What is the probability of picking a red or blue ball from a bag containing 3 red; 7 blue; and 5 green balls?

๐‘ƒ(๐‘… ๐‘œ๐‘Ÿ ๐ต) = ๐‘ƒ(๐‘…) + ๐‘ƒ(๐ต)

=3

15+

7

15

=2

3

Page 20: COUNTING & PROBABILITY

COMBINED EVENTS

oIf the combined events are NOT mutually exclusive, then

๐‘ƒ ๐ด ๐‘œ๐‘Ÿ ๐ต = ๐‘ƒ ๐ด + ๐‘ƒ ๐ต โˆ’ ๐‘ƒ(๐ด ๐‘Ž๐‘›๐‘‘ ๐ต)๐‘ƒ ๐ด โˆช ๐ต = ๐‘ƒ ๐ด + ๐‘ƒ ๐ต โˆ’ ๐‘ƒ(๐ด โˆฉ ๐ต)

o E.g. What is the probability of getting an even number or the number 4 when rolling a fair die?๐‘ƒ(๐ธ ๐‘œ๐‘Ÿ 4) = ๐‘ƒ(๐ธ) + ๐‘ƒ(4) โ€“ ๐‘ƒ(๐ธ ๐‘Ž๐‘›๐‘‘ 4)

=3

6+

1

6โˆ’

3

6ร—

1

4

=13

24

Page 21: COUNTING & PROBABILITY

Twelve numbers are dropped into a hat:

Event C = {๐‘“๐‘Ž๐‘๐‘ก๐‘œ๐‘Ÿ๐‘  ๐‘œ๐‘“ 6}

Event D = {๐‘“๐‘Ž๐‘๐‘ก๐‘œ๐‘Ÿ๐‘  ๐‘œ๐‘“ 9}

๐‘ƒ ๐ถ โˆช ๐ท = ๐‘ƒ ๐ถ + ๐‘ƒ ๐ท โˆ’ ๐‘ƒ(๐ถ โˆฉ ๐ท)

Combined Events Venn Diagram

Combined Events Card

Example

Combined Events Shape

& Colour Example

Page 22: COUNTING & PROBABILITY

Combined Events Question:

What is the probability of drawing a King or a Red card from a standard deck of cards?

๐‘ƒ(๐พ ๐‘œ๐‘Ÿ ๐‘…) = ๐‘ƒ(๐พ) + ๐‘ƒ(๐‘…) โ€“ ๐‘ƒ(๐พ ๐‘Ž๐‘›๐‘‘ ๐‘…)

=4

52+

26

52โˆ’

4

52ร—

26

52

=7

13

Page 23: COUNTING & PROBABILITY

Combined Events Question:

The probability of a wife going to the shops is 0,92; the probability of the husband going to the shops is 0,14; and the probability that both husband and wife will go to the shops is 0,45. Determine the probability of either the husband or the wife going to the shops.

๐‘ƒ(๐‘Š ๐‘œ๐‘Ÿ ๐ป) = ๐‘ƒ(๐‘Š) + ๐‘ƒ(๐ป) โ€“ ๐‘ƒ(๐‘Š ๐‘Ž๐‘›๐‘‘ ๐ป)

= 0,92 + 0,14 โˆ’ 0,45= 0,61 Combined Events

Questions

Page 24: COUNTING & PROBABILITY

COMPLEMENTARY EVENTS

o Given an event โ€œGโ€, a complementary event is an event that is not โ€œGโ€

o Complementary Rule: ๐‘ƒ ๐‘›๐‘œ๐‘ก ๐ด = 1 โˆ’ ๐‘ƒ ๐ด

๐‘ƒ ๐ดโ€ฒ = 1 โˆ’ ๐‘ƒ ๐ดo E.g. What is the probability of not

getting a 6 when rolling a fair die?๐‘ƒ 6โ€ฒ = 1 โˆ’ ๐‘ƒ 6

= 1 โˆ’1

6

=5

6

Page 25: COUNTING & PROBABILITY

Twelve numbers are dropped into a hat:

Event A = {๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ๐‘  โ‰ค 6}

Event B = {๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ๐‘  > 6}

๐‘ƒ ๐ดโ€ฒ = 1 โˆ’ ๐‘ƒ(๐ด)

Complementary Events Venn Diagram

Page 26: COUNTING & PROBABILITY

Complementary Events Question:

If you throw two dice, what is the probability of getting two different numbers?

๐‘ƒ ๐‘›๐‘œ๐‘ก ๐‘ ๐‘Ž๐‘š๐‘’ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ๐‘  = 1 โˆ’ (๐‘ ๐‘Ž๐‘š๐‘’ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ๐‘ )

= 1 โˆ’6

36

=5

6Rolling Doubles

Probability Laws Summary

Page 27: COUNTING & PROBABILITY

FUNDAMENTAL COUNTING PRINCIPLE

o If successive choices are made from ๐‘š1, ๐‘š2, ๐‘š3 , โ€ฆ ๐‘š๐‘› ; then the number of combined options is the product thereof:

๐‘š1 ร—๐‘š2 ร—๐‘š3 ร—โ‹ฏร—๐‘š๐‘›

o E.g. How many different outfits can I wear if I have 5 shirts, 3 pants and 4 shoes to wear?

5 ร— 3 ร— 4 = 60 ๐‘‘๐‘–๐‘“๐‘“๐‘’๐‘Ÿ๐‘’๐‘›๐‘ก ๐‘œ๐‘ข๐‘ก๐‘“๐‘–๐‘ก๐‘ 

Page 28: COUNTING & PROBABILITY

How many different combinations of meals could I order if I could choose from the following:

a soda or juice

a sandwich on a bagel, rye bread or white bread

with cheese, pastrami, roast beef or turkey as a filling?

Fundamental Counting Principle illustrated by a Tree Diagram

Page 29: COUNTING & PROBABILITY

2 ร— 3 ร— 4

= 24

Different meals

Fundamental Counting Principle illustrated by a Tree Diagram

Ways to Pick Officers Example

Page 30: COUNTING & PROBABILITY

Fundamental Counting Principle Question:

How many different party packs can be chosen from each of the following: BarOne; Kit-Kat; Smarties; Tex; Aero;

Nosh; or M&Mโ€™s Coke; Fanta; Cream Soda or Sprite Nik-Naks; Lays; or Simba Peanut butter or Marmite sandwich

7 ร— 4 ร— 3 ร— 2 = 168 ๐‘‘๐‘–๐‘“๐‘“๐‘’๐‘Ÿ๐‘’๐‘›๐‘ก ๐‘๐‘Ž๐‘Ÿ๐‘ก๐‘ฆ ๐‘๐‘Ž๐‘๐‘˜๐‘ 

Colour Combinations Question

Page 31: COUNTING & PROBABILITY

Fundamental Counting Principle Question:

How many possible letter arrangements can be made from the word โ€œMATHSโ€, if:

a) the letters may be repeated?

5 ร— 5 ร— 5 ร— 5 ร— 5 = 55 = 3125

b) the letters may NOT be repeated?

5 ร— 4 ร— 3 ร— 2 ร— 1 = 5! = 120

FACTORIAL NOTATION

Probability Summary Examples