MATH104 Ch. 11: Probability Theory part 3

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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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### MATH104Ch. 11: Probability Theorypart 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
• 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 1,6

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
• 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
• Note: These all sum to 1.
• Questions:
• Outcomes:
• TT, TH, HT, HH

1 2

H HH

H

T HT

T H TH

T TT

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

Note: sum of these outcomes is 1

Three Coins
• Which are most likely and why?
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
Theoretical Probabilities for 3 Coins
• Notice: Sum is 1.
Cards
• 4 suits, 13 denominations; 4*13=52 cards
• picture = J, Q, K
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)

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

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

…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

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

Find:

P(driver died)=

P(driver died/given no seat belt)=

P(no seat belt)=

P(no seat belt/given driver died)=

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 bday

What is E’ (not E)

Find P(E’)=