Math104 ch 11 probability theory part 3
This presentation is the property of its rightful owner.
Sponsored Links
1 / 41

MATH104 Ch. 11: Probability Theory part 3 PowerPoint PPT Presentation


  • 48 Views
  • Uploaded on
  • Presentation posted in: General

MATH104 Ch. 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. One Die.

Download Presentation

MATH104 Ch. 11: Probability Theory part 3

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


Math104 ch 11 probability theory part 3

MATH104Ch. 11: Probability Theorypart 3


Math104 ch 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


One die

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

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

Outcomes

  • For example, to get a sum of 6, you could get:

  • 5, 14,23,3…


Two dice theoretical probability

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

Sample Space for 2 Dice

1, 11, 21, 31, 41,51,6

2,12,22,32,42,52,6

3,13,23,33,43,53,6

4,14,24,34,44,54,6

5,15,25,35,45,55,6

6,16,26,36,46,56,6

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


Two dice team a b

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

Summary of 2 Dice (Team A/B) problem


Some probability rules and facts

Some Probability Rules and Facts

  • 0<= P(A) <= 1

  • Think of some examples where

    • P(A)=0P(A) = 1

  • The sum of all possible probabilities for an experiment is 1. Ex: P(Team A)+P(Team B) =1


One coin

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

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

Two Coins- Theoretical Answer

  • Outcomes:

  • TT, TH, HT, HH

    12

    HHH

    H

    THT

    THTH

    TTT


2 coins theoretical answer

2 Coins- Theoretical answer

P(0 heads) = 1/4

P(1 head, 1 tail)= 2/4 = 1/2

P(2 heads)= ¼

Note: sum of these outcomes is 1


Summary of 2 coin problem

Summary of 2 coin problem


Three coins

Three Coins

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

  • Which are most likely and why?


Three coins1

Three Coins

123

HHHHH

HTHHT

THHTHTHTT

THHTHH

TTHT

THTTH

2*2*2=8 outcomesTTTT


3 coins

3 coins

  • P(0 heads)=

  • P(1 head)=

  • P(2 heads)=

  • P(3 heads)=


Theoretical probabilities for 3 coins

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 problem

Summary of 3 coin problem


Cards

Cards

  • 4 suits, 13 denominations; 4*13=52 cards

  • picture = J, Q, K


When picking one card find

When picking one card, find…

  • P(heart)=

  • P(king)=

  • P(picture card)=

  • P(king or queen)=

  • P(king or heart)=


Theoretical probabilities cards

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

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

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 b1

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)


Math104 ch 11 probability theory part 3

Odds

Basic 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

Given probabilities, find odds

  1.      Recall probabilities

a.       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


Math104 ch 11 probability theory part 3

…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

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

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

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

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. 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

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

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

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

Conditional Probability

Find:

P(driver died)=

P(driver died/given no seat belt)=

P(no seat belt)=

P(no seat belt/given driver died)=


Math104 ch 11 probability theory part 3

  • 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

Birthday problem

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


Math104 ch 11 probability theory part 3

Hint

Let E=at least two people have the same bday

What is E’ (not E)

Find P(E’)=


  • Login