ch. 5: probability theory. probability assignment assignment by intuition – based on intuition,...
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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…
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”?
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?
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?