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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 .

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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
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, 1 4,2 3,3 …
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, 1 1, 2 1, 3 1, 4 1,5 1,6

2,1 2,2 2,3 2,4 2,5 2,6

3,1 3,2 3,3 3,4 3,5 3,6

4,1 4,2 4,3 4,4 4,5 4,6

5,1 5,2 5,3 5,4 5,5 5,6

6,1 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
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
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
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

1 2

H HH

H

T HT

T H TH

T TT

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

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

1 2 3

H H HHH

H T HHT

T H HTH T HTT

T H H THH

T THT

T H TTH

2*2*2=8 outcomes T TTT

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.
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)

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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

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…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)=

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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?

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Hint

Let E=at least two people have the same bday

What is E’ (not E)

Find P(E’)=

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