Ch. 5: Probability Theory. Probability Assignment Assignment by intuition – based on intuition, experience, or judgment. Assignment by relative frequency.

Ch. 5: Probability Theory

Probability Assignment

• Assignment by intuition – based on intuition, experience, or judgment.

• Assignment by relative frequency –

P(A) = Relative Frequency =

• Assignment for equally likely outcomes

n

f

Number of Outcomes Favorable to Event ( )

Total Number of Outcomes

AP A

One Die• Experimental Probability (Relative Frequency)

– If the class rolled one die 300 times and it came up a “4” 50 times, we’d say P(4)= 50/300

– The Law of Large numbers would say that our experimental results would approximate our theoretical answer.

• Theoretical Probability– Sample Space (outcomes): 1, 2, 3, 4, 5, 6– P(4) = 1/6– P(even) = 3/6

Two Dice

• Experimental Probability– “Team A” problem on the experiment: If we rolled

a sum of “6, 7, 8, or 9” 122 times out of 218 attempts, P(6,7,8, or 9)= 122/218= 56%

– Questions: What sums are possible?– Were all sums equally likely?– Which sums were most likely and why?– Use this to develop a theoretical probability– List some ways you could get a sum of 6…

Outcomes

• For example, to get a sum of 6, you could get:• 5, 1 4,2 3,3 …

Two Dice – Theoretical Probability

• Each die has 6 sides.• How many outcomes are there for 2 sides?

(Example: “1, 1”)• Should we count “4,2” and “2,4” separately?

Sample Space for 2 Dice

1, 1 1, 2 1, 3 1, 4 1,5 1,62,1 2,2 2,3 2,4 2,5 2,63,1 3,2 3,3 3,4 3,5 3,64,1 4,2 4,3 4,4 4,5 4,65,1 5,2 5,3 5,4 5,5 5,66,1 6,2 6,3 6,4 6,5 6,6

If Team A= 6, 7, 8, 9, find P(Team A)

Two Dice- Team A/B

• P(Team A)= 20/36• P(Team B) = 1 – 20/36 = 16/36• Notice that P(Team A)+P(Team B) = 1

Some Probability Rules and Facts

• 0<= P(A) <= 1• Think of some examples where

– P(A)=0 P(A) = 1• The sum of all possible probabilities for an

experiment is 1. Ex: P(Team A)+P(Team B) =1

One Coin

• Experimental– If you tossed one coin 1000 times, and 505 times

came up heads, you’d say P(H)= 505/1000– The Law of Large Numbers would say that this

fraction would approach the theoretical answer as n got larger.

• Theoretical– Since there are only 2 equally likely outcomes,

P(H)= 1/2

Two Coins

• Experimental Results– P(0 heads) = – P(1 head, 1 tail)=– P(2 heads)=– Note: These all sum to 1.

• Questions:– Why is “1 head” more likely than “2 heads”?

Two Coins- Theoretical Answer

• Outcomes: • TT, TH, HT, HH

1 2H HH

HT HT

T H THT TT

2 Coins- Theoretical answer

P(0 heads) = 1/4P(1 head, 1 tail)= 2/4 = 1/2P(2 heads)= ¼

Note: sum of these outcomes is 1

Three Coins

• Are “1 head” , “2 heads”, and “3 heads” all equally likely?

• Which are most likely and why?

Three Coins1 2 3

H H HHH H T HHT

T H HTHT HTT

T H H THHT THT

T H TTH2*2*2=8 outcomes T TTT

3 coins

• P(0 heads)=• P(1 head)= • P(2 heads)=• P(3 heads)=

Theoretical Probabilities for 3 Coins

• P(0 heads)= 1/8• P(1 head)= 3/8• P(2 heads)= 3/8• P(3 heads)= 1/8

• Notice: Sum is 1.

Cards• 4 suits, 13 denominations; 4*13=52 cards• picture = J, Q, K

A 2 3 4 5 6 7 8 9 10 J Q KHeart (red)

Diamond (red)Clubs (black)Spades (black)

When picking one card, find…

• P(heart)=• P(king)=• P(picture card)=• P(king or queen)=• P(king or heart)=

Theoretical Probabilities- Cards

• P(heart)= 13/52 = ¼ = 0.25• P(king)= 4/52= 1/13• P(picture card)= 12/52 = 3/13• P(king or queen)= 4/52 + 4 /52 = 8/52• P(king or heart)= 4/52 + 13/52 – 1/52 = 16/52

P(A or B)

• If A and B are mutually exclusive (can’t happen together, as in the king/queen example), then P(A or B)=P(A) + P(B)

• If A and B are NOT mutually exclusive (can happen together, as in the king/heart example), P(A or B)=P(A) + P(B) –P(A and B)

•

P (A and B)

• For independent events: P(A and B)• P(A and B) = P(A) * P(B)

• In General:• P(A and B) = P(A) * P(B/given A)

2 cards (independent) -questions

• Example: Pick two cards, WITH replacement from a deck of cards,

• P(king and king)=• P(2 hearts) =

P(A and B) Example-- Independent• For independent events: P(A and B)• P(A and B) = P(A) * P(B)• Example: Pick two cards, WITH replacement

from a deck of cards, • P(king and king)= 4/52 * 4/52 = 16/2704

=.0059• P(2 hearts) = 13/52 * 13/52 = .0625

P(A and B) – Dependent (without replacement)

• In General:• P(A and B) = P(A) * P(B/given A)• Example: Pick two cards, WITHOUT

replacement from a deck of cards, • P(king and king)= 4/52 * 3/51 =

12/2652=.0045• P(heart and heart)= 13/52 * 12/51 = 156/2652

= .059• P(king and queen) = 4/52 * 4/51 = 16/2652

Conditional Probability

Wore seat belt

No seat belt Total

Driver survived

412,368 162,527 574,895

Driver died 510 1601 2111

Total 412,878 164,128 577,006

Find: P(driver died)=P(driver died/given no seat belt)=P(no seat belt)= P(no seat belt/given driver died)=

Wore seat belt

No seat belt

Total

Driver survived

412,368 162,527 574,895

Driver died

510 1601 2111

Total 412,878 164,128 577,006

• P(driver died)= 2111/577,006 = .00366• P(driver died/given no seat belt)= 1601/164,128

= .0097• P(no seat belt)= 164,128/577,006= .028• P(no seat belt/given driver died)= 1602/2111= .76

Multiplication Problems• 1. At a restaurant, you have a choice of main dish (beef, chicken, fish,

vegetarian), vegetable (broccoli, corn), potato (baked, fries), and dessert (chocolate, strawberry). LIST all possible choices.

• • 2. A teacher wishes to make all possible different answer keys to a

multiple choice quiz. How many possible different answer keys could there be if there are 3 questions that each have 4 choices (A,B,C,D). LIST them all.

• • 3. What if there were 20 multiple choice questions with 5 choices each?

Explain (don’t list).• • 4. With 9 baseball players on a team, how many different batting orders

exist?

Answers

• 1. At a restaurant, you have a choice of main dish (beef, chicken, fish, vegetarian), vegetable (broccoli, corn), potato (baked, fries), and dessert (chocolate, strawberry). LIST all possible choices.

• main vegetable potato dessert

– Beef broc baked chocolate– Beef broc baked strawb– Beef broc fries chocolate– …

–4*2*2*2=32

Answers

• 2. A teacher wishes to make all possible different answer keys to a multiple choice quiz. How many possible different answer keys could there be if there are 3 questions that each have 4 choices (A,B,C,D). LIST them all. 4*4*4=64

• • 3. What if there were 20 multiple choice questions with

5 choices each? Explain (don’t list). 5^20• • 4. With 9 baseball players on a team, how many

different batting orders exist? 9! = 362,880

Permutation Examples

1. If there are 4 people in the math club (Anne, Bob, Cindy, Dave), and we wish to elect a president and vice-president, LIST all of the different ways that this is possible.

2. From these 4 people (Anne, Bob, Cindy, Dave),

we wish to elect a president, vice-president, and treasurer. LIST all of the different ways that this is possible.

Answers

1. If there are 4 people in the math club (Anne, Bob, Cindy, Dave), and we wish to elect a president and vice-president, LIST all of the different ways that this is possible.

AB BA CA DAAC BC CB DBAD BD CD DC

4*3=12 or 4P2 = 12

Answers2. From these 4 people (Anne, Bob, Cindy, Dave),

we wish to elect a president, vice-president, and treasurer. LIST all of the different ways that this is possible.

ABCABD…

• A B C ABC

D ABDC B ACB

D ACDD A BDA

C BDC• B A C BAC

D BCDC A BCA

D BCDD A BDA

C BDC• C A B CAB

D CADB A CBA

D CBDA B DAB

C DAC• D A B DAB

C DACB A DBA

C DBCC A DCA

B DCB

4*3*2 = 24 outcomesOr 4P3 = 24

Combination Examples

1. If there are 4 people in the math club (Anne, Bob, Cindy, Dave), and 2 will be selected to attend the national math conference. LIST all of the different ways that this is possible.

2. From these 4 people (Anne, Bob, Cindy, Dave), and 3 will be selected to attend the national math conference. LIST all of the different ways that this is possible.

Combination answers1. If there are 4 people in the math club (Anne,

Bob, Cindy, Dave), and 2 will be selected to attend the national math conference. LIST all of the different ways that this is possible.

ABAC BCAD BD CD

4C2= 6

Combination answer

2. From these 4 people (Anne, Bob, Cindy, Dave), and 3 will be selected to attend the national math conference. LIST all of the different ways that this is possible.

ABC BCDABDACD

4C3 = 4

Permutations and Combinations• Permutations

– Use when ORDER matters and NO repitition– nPr = n!/(n-r)!– Example: If 10 people join a club, how many ways

could we pick pres and vp? 10P2 = 90• Combinations

– Use: ORDER does NOT matter and NO repitition– nCr = n!/ [(n-r)!r!]– Example: 10 people join a club. In how many ways

could we pick 2? 10C2 = 45

Multiplication, Permutation, or Combination?

1. With 14 players on a team, how many ways could we pick a batting order of 11?

2. If license plates have 3 letters and then 4 numbers, how many

different license plates exist? 3. How many different four-letter radio station call letters can be formed

if the first letter must be W or K? 4. A social security number contains nine digits. How many different

ones can be formed? 5. If you wish to arrange your 7 favorite books on a shelf, how many

different ways can this be done?

6. If you have 10 favorite books, but only have room for 7 books on the shelf, how many ways can you arrange them?

7. You wish to arrange 12 of your favorite photographs on a mantel. How many ways can this be done?

8. You have 20 favorite photographs and wish to arrange 12 of them on a mantel. How many ways can that be done?

9. You take a multiple choice test with 12 questions (and each can be answered A B C D E). How many different ways could you answer the test?

10. If you had 13 pizza toppings, how many ways could you pick 5 of them?

Answers

1. 14P11 =175,760,000 6. 10P7

2. 26*26*26*10*10*10*10 7. 12! or 12P12

3. 2*26*26*26 8. 20P12

4. 10^9 9. 5^12

5. 7! Or 7P7 10. 13 C5