MATH104Ch. 11: Probability Theory

part 3

Probability Assignment

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

• Assignment by relative frequency – P(A) = Relative Frequency =

• Assignment for equally likely outcomes

nf

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

Summary of 2 Dice (Team A/B) problem

TOSSING 1 DIE P(1) P(3) P(5) P(even number)

Sample Empirical ResultsUse P(E)= n(E) / total tosses

5/30 = .17 6/30 = .2 4/30 = .13 15/30 = .5

Your Empirical Results

Class Empirical Results 51/300 = .17 48/ 300 = .16 53/ 300 = .18 148/300

= .49Theoretical ResultsUse P(E)= n(E) / n(S)where S=SampleSpace

1/6 =.167 1/6 =.167 1/6 =.167 3/6 = 1/2

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

Summary of 2 coin problemTOSSING 2 COINS P(2 heads, 0 tails) P(1 head, 1 tail) P(0 heads, 2 tails)Guess: how often would you expect to get each possibility?

Sample Empirical ResultsUse P(E)= n(E) / total tosses

8/30 = .27 14/30 = .47 8/30 = .27

Your Empirical Results

Class Empirical Results 73/ 300 = .243 151/ 300 = .5 76/ 300 = .253

Theoretical Results

Use P(E)= n(E) / n(S)1 /4 2/4 = 1/2 1 /4

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.

Summary of 3 coin problemTOSSING 3 COINS P(3 heads) P(2 heads) P(1 head) P(0 heads)

Sample Empirical ResultsUse P(E)= n(E) / total tosses

3/20= .15 7/20 = .35 6/20 = .3 4/20 = .2

Your Empirical Results

Class Empirical Results 26/ 200 = .13 74 / 200 = .37 76 / 200 = .38 24/ 200 = .12

Theoretical Results

Use P(E)= n(E) / n(S)

…

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

11.6 Not, Mutually Exclusive, Odds

P(E)= 1-P(E ‘ ) where E’ = not E=complement of E 1. If there is a 20% chance of snow tomorrow, what is the

chance it will not snow tomorrow?

2. When choosing one card from a deck, find the probability of selecting:

a. A heart b. A card that is not a heart c. A king d. A card that is not a king

P(A or B)1. When selecting one card, find the probability of:a. king or queen b. king or a heart c. king or a 5

d. 5 or a diamond

e. Picture card or a 7

f. Picture card or a red card

P(A or B)

• Mutually exclusive events—cannot occur together

• If A and B are mutually exclusive, P(A or B) = P(A) + P(B)

• If A and B are not mutually exclusive, P(A or B) = P(A) + P(B) – P(A and B)

OddsBasic idea: If, when drawing one card from a deck, the

probability of getting a heart is ¼, then The odds in favor of drawing a heart are 1:3 and the odds against

a heart are 3:1. Another example: If, when drawing one card from a deck, the

probability of getting a king is 1/13, then The odds in favor of drawing a king are 1:12, and the odds

against a king are 12:1.

Odds to Probability if odds in favor of E are a:b, then P(E)=

Given probabilities, find odds 1. Recall probabilitiesa. P(heart) b. P(not a heart) c. P(king) d. P(picture card) e. P(red card)

1. Find the odds in favor of:a. A heart b. A card that is not a heart c. A king d. A picture card e. A red card

…odds

2. If the odds in favor of winning the lottery are 1:1,000,000, find the probability of winning the lottery

3. If the odds in favor of getting a certain job are

3:4, find the probability of getting the job.

11.7: And, Conditional

Independent events- two events are independent events if the occurrence of either of them has no effect on the probability of the other

If A and B are independent events, then P(A and

B) = P(A)*P(B)

2 kids

1. Assuming it’s equally likely that boys and girls are born, in a family with 2 kids, find the probability of getting:

a. 2 girls b. 2 boys

2. In a family with 3 kids, find the probability of getting:

Assuming P(B)=P(G)a. 3 girls b. 3 boys c. At least 1 boy

3. In a family of 4 kids, find the probability of getting:

a. 4 girls b. 4 boys c. At least 1 boy

4. Two cards

4. If you pick two cards out of a deck of cards and replace them in between picks, find:

a. P( 2 red cards) b. P(2 hearts)

c. P(2 kings)

Dependent events—

the occurrence of one of them has an effect on the occurrence of the other

If A and B are dependent, P(A and B) = P(A)*P(B,

given A)

Without replacement:

1. If you pick two cards out a deck without replacement, find the probability of getting:

a. 2 red cards b. 2 kings

2. pick 3 cards without replacement

find the probability of getting: a. 3 red cards b. 3 kings c. A king, then a queen, then a jack (in that

order)

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

Birthday problem

What is the probability that two people in this class would have the same birth date?

Hint

Let E=at least two people have the same bdayWhat is E’ (not E)

Find P(E’)=