MATH 20: PROBABILITY Conditional Probability Discrete Co Xingru - - PowerPoint PPT Presentation
MATH 20: PROBABILITY Conditional Probability Discrete Co Xingru Chen xingru.chen.gr@dartmouth.edu Monty Hall Problem 03 03 Suppose you're on a game show, and you're given the choice of three doors: behind one door is a
Discrete Co Conditional Probability Xingru Chen xingru.chen.gr@dartmouth.edu
§ Suppose you're
a game show, and you're given the choice
three doors: behind
door is a car; behind the
goats.
§ You pick a door, say
and the host, who knows what's behind the doors,
another door, say
which has a goat.
§ He then says to you, "Do you want to pick door
§ Is it to your advantage to switch your choice?
How many possible arrangements
car and two goats behind three doors?
Find where the car is:
! " = 3.
What is your probability
winning the car after you pick door No.1?
What is your probability
winning if you switch?
What is your probability
winning if you stick to your initial choice?
What is your probability
winning if you switch?
What is your probability
winning if you stick to your initial choice? Be Behind door 1 Be Behind door 2 Be Behind door 3 St Staying at door 1 Sw Switching car goat goat win car win goat goat car goat win goat win car goat goat car win goat win car
Be Behind nd doors rs St Stay aying Sw Switching car goat goat win car win goat goat car goat win goat win car goat goat car win goat win car probability
winning 𝟐 𝟒 𝟑 𝟒
there is more in informatio ion about doors 2 and 3 than was available at the beginning
the game!
Blindly pick up a ball.
Ex Exper erimen ent
! "# = " $
# "# = " %
% "# = " #
Outco come 𝒀
! "# = " $
# "# = " %
% "# = " #
Outco come 𝒀
𝑄 𝐹 = 1 2
Even Event 𝑭: : the ball is blue
green
𝑄 𝐹 = 2 3
Even Event 𝑮: : the ball is brown
green
! "# = " $
# "# = " %
% "# = " #
Outco come 𝒀
𝑄 𝐹 = 1 2
Even Event 𝑭: : the ball is blue
green
𝑄 𝐹 = 2 3
Even Event 𝑮: : the ball is brown
green
The ball is picked up and we are told that Event 𝐹 has
𝑄 𝐹|𝐺 = ⋯
Even Event 𝑮 gi given en 𝑭: the ball is brown
green kn known tha that the the ball is blue
green
! "# = " $
# "# = " %
% "# = " #
Outco come 𝒀
𝑄 𝐹 = 1 2
Even Event 𝑭: : the ball is blue
green
𝑄 𝐹 = 2 3
Even Event 𝑮: : the ball is brown
green
The ball is picked up and we are told that Event 𝐹 has
𝑄 𝐹|𝐺 = 2 6 = 1 3
Even Event 𝑮 gi given en 𝑭: : the ball is brown
green kn known that the ball is blue
green
there is more in informatio ion about the event than was available at the beginning
the experiment.
How to compute 𝑄(𝐺|𝐹)?
§ Let Ω = 𝜕!, 𝜕", ⋯ , 𝜕# be the
sample space with distribution function 𝑛(𝜕$) assigned. § Given that the event 𝐹 has
(𝑄(𝐹) > 0), the conditional probability 𝑛 𝜕% 𝐹 = 𝑑𝑛(𝜕%) for all 𝜕% in 𝐹. ∑& 𝑛 𝜕% 𝐹 = 𝑑 ∑& 𝑛(𝜕%) = 𝑑𝑄 𝐹 = 1 ⟹ 𝑑 =
! '(&).
§ Thus, we define the conditional distribution given 𝐹 𝑛 𝜕% 𝐹 = *(+!)
'(&) .
§ Let Ω = 𝜕!, 𝜕", ⋯ , 𝜕# be the
sample space with distribution function 𝑛(𝜕$) assigned. § Given that the event 𝐹 has
(𝑄(𝐹) > 0), the conditional probability 𝑛 𝜕% 𝐹 = ⋯
𝑛 𝜕# 𝐹 = 𝑛(𝜕#) 𝑄(𝐹)
𝑛 𝜕# 𝐹 = ⋯
Rolling a die once.
Ex Exper erimen ent
Ω = 1, 2, 3, 4, 5, 6
Sample Space ce 𝛁
𝑄 𝐹 = 2 6 = 1 3
Even Event 𝑭: : 𝒀 > 𝟓
𝑛 1 = 𝑛 2 = 𝑛 3 = 𝑛 4 = 𝑛 5 = 𝑛 6 = 1 6
Distribution Funct ction 𝒏(𝝏𝒌)
Ω = 1, 2, 3, 4, 5, 6
Sample Space ce 𝛁
𝑄 𝐹 = 2 6 = 1 3
Even Event 𝑭: : 𝒀 > 𝟓
𝑛 1 = 𝑛 2 = 𝑛 3 = 𝑛 4 = 𝑛 5 = 𝑛 6 = 1 6
Distribution Funct ction 𝒏(𝝏𝒌)
Sa Sample Su Sub-Space ce
"/! = " '
Co Conditional Proba babi bility
𝑛(𝜕&) = 1 6
Distribution Funct ction 𝒏(𝝏𝒌)
𝑄 𝐹 = 2 6 = 1 3
Even Event 𝑭: : 𝒀 > 𝟓
The condition probability
a general event 𝐺 given that 𝐹
𝑄 𝐺 𝐹 = ∑(∩* 𝑛 𝜕# 𝐹 = ∑(∩*
+(-!) /(*) = /((∩*) /(*) .
1(4) .
𝜕# in 𝐹 𝑛 𝜕# 𝐹 = 𝑛(𝜕#) 𝑄(𝐹) 𝜕# not in 𝐹 𝑛 𝜕# 𝐹 = 0
𝑄 𝐺 𝐹 = 1/6 1/3 = 1 2
Co Conditional Proba babi bility
𝑄 𝐹 = 2 6 = 1 3
Even Event 𝑭: : 𝒀 > 𝟓
𝑄 𝐺 = 3 6 = 1 2
Even Event 𝑮: : 𝒀 is is even
1(4) . 𝑄 𝐹 ∩ 𝐺 = 1 6
Even Event 𝑭 ∩ 𝑮: : 𝒀 = 𝟕
𝑄 𝐺 𝐹 = 0 2 = 1/3 = 0
Co Conditional Proba babi bility
𝑄 𝐹 = 2 6 = 1 3
Even Event 𝑭: : 𝒀 > 𝟓
𝑄 𝐺 = 2 6 = 1 3
Even Event 𝑮: : 𝒀 is is a sq square number
1(4) . 𝑄 𝐹 ∩ 𝐺 = 0
Even Event 𝑭 ∩ 𝑮: : ∅
§ Suppose you're
a game show, and you're given the choice
three doors: behind
door is a car; behind the
goats.
boys given that it has at least
boy?
boys given that the first child is a boy?
girl given that it has at least
boy?
girl given that the first child is a boy?
§ Suppose you're
a game show, and you're given the choice
three doors: behind
door is a car; behind the
goats.
boys given that it has at least
boy?
boys given that the first child is a boy?
girl given that it has at least
boy?
girl given that the first child is a boy?
1(4) .
two boys
girl
at least
boy the first child is a boy
Ω = 𝐶𝐶, 𝐶𝐻, 𝐻𝐶, 𝐻𝐻
Sample Space ce 𝛁
𝑄 = ⋯
Even Event: at leas east
boy
𝑛 𝐶𝐶 = 𝑛 𝐶𝐻 = 𝑛 𝐻𝐶 = 𝑛 𝐻𝐻 = 1 4
Distribution Funct ction 𝒏(𝝏𝒌)
A family of two children.
Ex Exper erimen ent 1 1 2 1 3
𝑄 = ⋯
Event: the first ch child is a boy
𝑄 = ⋯
Even Event: two boys
𝑄 = ⋯
Even Event:
gi girl
𝑄 𝐹 = 1 − 1 4 = 3 4
Even Event 𝑭: : at least
boy 1 1 2 1 3 What is the probability that a family
two children has
boys given that it has at least
boy?
1(4) . 𝑄 𝐺 = 1 4
Even Event 𝑮: : two boys
𝑄 𝐹 ∩ 𝐺 = 1 4
Even Event 𝑭 ∩ 𝑮: : two boys
𝑄 𝐺 𝐹 =
'()∩+) '(+) = "/! $/! = " $.
𝑄 𝐹 = 1 2
Even Event 𝑭: : the first is a boy 1 1 2 1 3 What is the probability that a family
two children has
boys given that the first child is a boy?
1(4) . 𝑄 𝐺 = 1 4
Even Event 𝑮: : two boys
𝑄 𝐹 ∩ 𝐺 = 1 4
Even Event 𝑭 ∩ 𝑮: : two boys
𝑄 𝐺 𝐹 =
'()∩+) '(+) = "/! "/# = " #.
𝑄 𝐹 = 1 − 1 4 = 3 4
Even Event 𝑭: : at least
boy 1 1 2 1 3 What is the probability that a family
two children has
girl given that it has at least
boy?
1(4) . 𝑄 𝐺 = 1 2
Even Event 𝑮: :
girl
𝑄 𝐹 ∩ 𝐺 = 1 2
Even Event 𝑭 ∩ 𝑮: :
boy and
girl
𝑄 𝐺 𝐹 =
'()∩+) '(+) = "/# $/! = # $.
𝑄 𝐹 = 1 2
Even Event 𝑭: : the first is a boy 1 1 2 1 3 What is the probability that a family
two children has
girl given that the first child is a boy?
1(4) . 𝑄 𝐺 = 1 2
Even Event 𝑮: :
girl
𝑄 𝐹 ∩ 𝐺 = 1 4
Even Event 𝑭 ∩ 𝑮: : the first is a boy and the seco cond is a gi girl
𝑄 𝐺 𝐹 =
'()∩+) '(+) = "/! "/# = " #.
§ Suppose you're
a game show, and you're given the choice
three doors: behind
door is a car; behind the
goats. What is the probability that a family
two children has
boys given that it has at least
boy? 𝑄 𝐺 = "
0,
𝑄 𝐺 𝐹 = "
!.
boys given that the first child is a boy? 𝑄 𝐺 = "
0,
𝑄 𝐺 𝐹 = "
'.
girl given that it has at least
boy? 𝑄 𝐺 = "
',
𝑄 𝐺 𝐹 = '
!.
girl given that the first child is a boy? 𝑄 𝐺 = "
',
𝑄 𝐺 𝐹 = "
'.
Tow events 𝐹 and 𝐺 are independent if both 𝐹 and 𝐺 have positive probability and if 𝑄(𝐹|𝐺) = 𝑄(𝐹), and 𝑄(𝐺|𝐹) = 𝑄(𝐺). What is the probability that a family
two children has
girl given that the first child is a boy? 𝑄 𝐺 = "
',
𝑄 𝐺 𝐹 = "
'.
§ Suppose you're
a game show, and you're given the choice
three doors: behind
door is a car; behind the
goats.
A A Philosophical Essay
Probabilities I have seen men, ardently desirous
having a son, who could learn
with anxiety
the births
boys in the month when they expected to become fathers. Imagining that the ratio
these births to those
girls
to be the same at the end
each month, they judged that th the boys already born rn wo would render more probable the births next
gi girls. Pi Pierre-Si Simon Lapla place
§ Suppose you're
a game show, and you're given the choice
three doors: behind
door is a car; behind the
goats.
Mo Monte Carlo Casin ino in a game
the Monte Carlo Casino on August 18, 1913, the ball fell in black 26 times in a row. This was an extremely uncommon
the probability
a sequence
either red
black
26 times in a row is around 1 in 66.6 million, assuming the mechanism is unbiased. Ga Gamblers lost millions
francs betting against black, k, reasoning incorrectly that th the str treak was causing an imbalance in th the randomness
th the wheel, and th that it ha had to be fol
by a lon
streak
re red.
§ Suppose you're
a game show, and you're given the choice
three doors: behind
door is a car; behind the
goats.
The ga gambler er's fallacy cy, also known as the Mo Monte Carlo fallacy cy or the fallacy
the maturity
chances, is the erroneous belief that if if a particu icular event
ccurs more frequ quently than normal during the past it is less like kely to happen in the future (or vice versa), when it has
been established that the probability
such events does not depend
what has happened in the past. Such events, having the quality
historical independence, are referred to as st statist stically lly in
commonly associated with gambling, where it may be believed, for example, that the next dice roll is more than usually likely to be six because there have recently been less than the usual number
sixes.
Tow events 𝐹 and 𝐺 are independent if both 𝐹 and 𝐺 have positive probability and if 𝑄(𝐹|𝐺) = 𝑄(𝐹), and 𝑄(𝐺|𝐹) = 𝑄(𝐺).
§ Suppose you're
a game show, and you're given the choice
three doors: behind
door is a car; behind the
goats.
Tow events 𝐹 and 𝐺 are independent if both 𝐹 and 𝐺 have positive probability and if 𝑄(𝐹|𝐺) = 𝑄(𝐹), and 𝑄(𝐺|𝐹) = 𝑄(𝐺). If 𝑄 𝐹 > 0 and 𝑄 𝐺 > 0, the 𝐹 and 𝐺 are independent if and
if 𝑄 𝐹 ∩ 𝐺 = 𝑄 𝐹 𝑄(𝐺).
𝑄 𝐹 𝐺 = /(*∩()
/(() .
𝑄 𝐺 𝐹 = /((∩*)
/(*) .
If 𝑄 𝐹 > 0 and 𝑄 𝐺 > 0, the 𝐹 and 𝐺 are independent if and
if 𝑄 𝐹 ∩ 𝐺 = 𝑄 𝐹 𝑄(𝐺).
A set
events 𝐵", 𝐵', ⋯ , 𝐵1 is said to be mutually independent if for any subset 𝐵2, 𝐵3, ⋯ , 𝐵+ of these events we have 𝑄 𝐵2 ∩ 𝐵3 ∩ ⋯ 𝐵+ = 𝑄 𝐵2 𝑄(𝐵3) ⋯ 𝑄(𝐵+). Or equivalently, if for any sequence ̅ 𝐵", ̅ 𝐵', ⋯ , ̅ 𝐵1, with ̅ 𝐵2 = 𝐵2 or ̅ 𝐵2 = A 𝐵2, 𝑄 ̅ 𝐵" ∩ ̅ 𝐵' ∩ ⋯ ̅ 𝐵1 = 𝑄 ̅ 𝐵" 𝑄( ̅ 𝐵') ⋯ 𝑄( ̅ 𝐵1).
Even Event-1: : rain 𝑄 rain = 0.6 Even Event-2: windy & cl cloudy 𝑄 windy & cloudy = 0.48 Even Event-3:
people ca carrying umbrellas 𝑄 umbrella = 0.56 Even Event-4: 4: the the denti tist is
to today 𝑄 dentist = 5 7
Even Event-1: 1: rain 𝑄 rain = 0.6 Even Event-2: windy & cl cloudy 𝑄 windy & cloudy = 0.48 𝑄 rain | windy & cloudy = 0.85 Even Event-3:
people ca carrying umbrellas 𝑄 umbrellas = 0.56 𝑄 rain | umbrellas = 0.9 Even Event-4: 4: the the denti tist is
to today
𝑄 dentist = 5 7 𝑄 rain | dentist = 0.6
G, 𝑌H = 𝑠H, ⋯ , 𝑌I = 𝑠 I = 𝑄 𝑌G = 𝑠 G 𝑄(𝑌H = 𝑠H) ⋯ 𝑄(𝑌I = 𝑠 I),
G, 𝑠H, ⋯ , 𝑠 I.
relation probability statistics
In a group
60 people, the numbers who do
do not smoke and do
do not have cancer are reported as shown below. Cancer Smoke
Not Smoke ke Smoke ke To Total Not Cancer 40 10 50 Cancer 7 3 10 Total 47 13 60
Let Ω be the sample space consisting
these 60
person is chosen at random from the
𝐷(𝜕) = 1 if this person has cancer and 0 if not, and 𝑇(𝜕) = 1 if this person smokes and 0 if not.
Ca Cancer
Variable: 𝐷
Space: 0, 1
Function: 𝑄 𝐷 = 0 = 50 60 = 5 6 𝑄 𝐷 = 1 = 10 60 = 1 6 Smoke ke
Variable: 𝑇
Space: 0, 1
Function: 𝑄 𝑇 = 0 = 47 60 𝑄 𝑇 = 1 = 13 60
Cancer Smoke the ma marginal distributions of 𝐷 and 𝑇
Cance cer and Smoke ke
Variable: 𝑌 = (𝐷, 𝑇)
Space: Ω = 0, 1 × 0, 1
Distribution Function: Cancer Smoke the jo joint nt distribution of {𝐷, 𝑇} 𝑇
1
𝐷
40/60 10/60 1 7/60 3/60
Cancer Smoke the jo joint nt distribution of {𝐷, 𝑇} 𝑇
1
𝐷
40/60 10/60 1 7/60 3/60
Ca Cancer
Variable: 𝐷
Space: 0, 1
Function: 𝑄 𝐷 = 0 = 50 60 = 5 6 𝑄 𝐷 = 1 = 10 60 = 1 6 Smoke ke
Variable: 𝑇
Space: 0, 1
Function: 𝑄 𝑇 = 0 = 47 60 𝑄 𝑇 = 1 = 13 60
the ma marginal distributions of 𝐷 and 𝑇
the joint distribution
{𝐷, 𝑇}
𝑇 1 𝐷 40/60 10/60 1 7/60 3/60 Ca Cancer
Variable: 𝐷
Space: 0, 1
Function: 𝑄 𝐷 = 0 = 50 60 = 5 6 𝑄 𝐷 = 1 = 10 60 = 1 6 Smoke ke
Variable: 𝑇
Space: 0, 1
Function: 𝑄 𝑇 = 0 = 47 60 𝑄 𝑇 = 1 = 13 60
the marginal distributions
𝐷 and 𝑇
Tow events 𝐹 and 𝐺 are independent if both 𝐹 and 𝐺 have positive probability and if 𝑄(𝐹|𝐺) = 𝑄(𝐹), and 𝑄(𝐺|𝐹) = 𝑄(𝐺).
Are 𝐷 and 𝑇 independent?
𝑄 𝐷 = 1 𝑇 = 1 = ⋯ 𝑄 𝑇 = 1 𝐷 = 1 = ⋯
the joint distribution of {𝐷, 𝑇}
𝑇 1 𝐷 40/60 10/60 1 7/60 3/60 Ca Cancer
Variable: 𝐷
Space: 0, 1
Function: 𝑄 𝐷 = 0 = 50 60 = 5 6 𝑄 𝐷 = 1 = 10 60 = 1 6 Smoke ke
Variable: 𝑇
Space: 0, 1
Function: 𝑄 𝑇 = 0 = 47 60 𝑄 𝑇 = 1 = 13 60
The marginal distributions of 𝐷 and 𝑇
Tow events 𝐹 and 𝐺 are independent if both 𝐹 and 𝐺 have positive probability and if 𝑄(𝐹|𝐺) = 𝑄(𝐹), and 𝑄(𝐺|𝐹) = 𝑄(𝐺).
Are 𝐷 and 𝑇 independent?
𝑄 𝐷 = 1 𝑇 = 1 = 3 13 𝑄 𝑇 = 1 𝐷 = 1 = 3 10
the joint distribution of {𝐷, 𝑇}
𝑇 1 𝐷 40/60 10/60 1 7/60 3/60 Ca Cancer
Variable: 𝐷
Space: 0, 1
Function: 𝑄 𝐷 = 0 = 50 60 = 5 6 𝑄 𝐷 = 1 = 10 60 = 1 6 Smoke ke
Variable: 𝑇
Space: 0, 1
Function: 𝑄 𝑇 = 0 = 47 60 𝑄 𝑇 = 1 = 13 60
The marginal distributions of 𝐷 and 𝑇
If 𝑄 𝐹 > 0 and 𝑄 𝐺 > 0, the 𝐹 and 𝐺 are independent if and
if 𝑄 𝐹 ∩ 𝐺 = 𝑄 𝐹 𝑄(𝐺).
Are 𝐷 and 𝑇 independent?
𝑄(𝐷 = 1, 𝑇 = 1) = 3 60 𝑄 𝐷 = 1 𝑄 𝑇 = 1 = 1 6 13 60 = 13 360
mutually independent & same distribution
A sequence
random variables 𝑌", 𝑌', ⋯ , 𝑌1 that are mu mutually independent and that have the the same distr tributi tion is called a sequence
independent trials
independent trials process.
G, 𝑠H, ⋯ , 𝑠 M ,
O, O- ⋯ O. P, P- ⋯
𝑆 = 1, 2, 3, 4, 5, 6 𝑛4 = 1 2 3 4 5 6 1 6 1 6 1 6 1 6 1 6 1 6
Q denote