Transcript Ch 11 Power point- part 3 (probability notes)

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 =
f
n
• Assignment for equally likely outcomes
Number of Outcomes Favorable to Event A
P( A) 
Total Number of 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
2,1
3,1
4,1
5,1
6,1
1, 2
2,2
3,2
4,2
5,2
6,2
1, 3
2,3
3,3
4,3
5,3
6,3
1, 4
2,4
3,4
4,4
5,4
6,4
1,5
2,5
3,5
4,5
5,5
6,5
1,6
2,6
3,6
4,6
5,6
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
Results
Use P(E)=
n(E) / total tosses
5/30 = .17
6/30 = .2
4/30 = .13
15/30 = .5
Class Empirical
Results
51/300 = .17
48/ 300 = .16
53/ 300 = .18
148/300 =
.49
Theoretical Results
Use P(E)= n(E) / n(S)
where
S=SampleSpace
1/6 =.167
1/6 =.167
1/6 =.167
3/6 = 1/2
Your Empirical
Results
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
2
H
HH
T
HT
H
T
TH
TT
H
T
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
TOSSING 2 COINS
P(2 heads, 0 tails)
P(1 head, 1 tail)
P(0 heads, 2 tails)
8/30 = .27
14/30 = .47
8/30 = .27
73/ 300 = .243
151/ 300 = .5
76/ 300 = .253
1 /4
2/4 = 1/2
1 /4
Guess: how often
would you expect
to get each
possibility?
Sample Empirical
Results
Use P(E)=
n(E) / total tosses
Your Empirical
Results
Class Empirical
Results
Theoretical Results
Use P(E)= n(E) / n(S)
Three Coins
• Are “1 head” , “2 heads”, and “3 heads” all
equally likely?
• Which are most likely and why?
Three Coins
1
2
H
H
T
T
H
T
2*2*2=8 outcomes
3
H
T
H
T
H
T
H
T
HHH
HHT
HTH
HTT
THH
THT
TTH
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 problem
TOSSING 3
COINS
P(3 heads)
P(2 heads)
P(1 head)
P(0 heads)
Sample
Empirical
Results
Use P(E)=
n(E) / total
tosses
3/20= .15
7/20 = .35
6/20 = .3
4/20 = .2
26/ 200 = .13
74 / 200 = .37
76 / 200 = .38
24/ 200 = .12
Your Empirical
Results
Class Empirical
Results
Theoretical
Results
…
Use P(E)=
n(E) / n(S)
Cards
• 4 suits, 13 denominations; 4*13=52 cards
• picture = J, Q, K
A
Heart
(red)
Diamond
(red)
Clubs
(black)
Spades
(black)
2
3
4
5
6
7
8
9
10
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 ‘ )
1.
2.
where E’ = not E=complement of E
If there is a 20% chance of snow tomorrow, what is the
chance it will not snow tomorrow?
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.
a.
b.
c.
When selecting one card, find the probability of:
king or queen
king or a heart
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.
a.
Recall probabilities
P(heart)
1.
a.
Find the odds in favor of:
A heart
b.
P(not a heart)
b.
A card that is not a heart
c.
P(king)
c.
A king
d.
P(picture card)
d.
A picture card
e. P(red 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
No seat
seat belt belt
Total
Driver
412,368
survived
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 bday
What is E’ (not E)
Find P(E’)=