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 car; behind the 02 02 others, goats. 01 01

Monty Hall Problem § You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. 03 03 02 02 01 01

Monty Hall Problem § He then says to you, "Do you want to pick door No. 2?” § Is it to your advantage to switch your choice? 03 03 02 02 01 01

01 01 Monty Hall Problem 02 02 03 03 How many possible arrangements of one car and two goats behind three doors? Find where the car is: = ! " = 3 .

01 01 02 02 03 03 What is your probability of winning ? the car after you pick door No.1? What is your probability of winning if ? you stick to your initial choice? What is your probability of winning if ? you switch?

What is your probability of winning ? if you stick to your initial choice? What is your probability of winning ? if you switch? 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 of 𝟐 𝟑 winning 𝟒 𝟒 there is more in informatio ion about doors 2 and ! 3 than was available at the beginning of the game!

Colorful Balls in a Jar Ex Exper erimen ent Blindly pick up a ball. Outco come 𝒀 ! " • 𝑄 𝑌 = blue ball = "# = $ # " • 𝑄 𝑌 = green ball = "# = % % " 𝑄 𝑌 = brown ball = "# = • #

Colorful Balls in a Jar Outco come 𝒀 ! " 𝑄 𝑌 = blue ball = "# = • $ # " 𝑄 𝑌 = green ball = "# = • % % " 𝑄 𝑌 = brown ball = "# = • # Event Even 𝑭 : : the ball is blue or green 𝑄 𝐹 = 1 2 Even Event 𝑮 : : the ball is brown or green 𝑄 𝐹 = 2 3

Event Even 𝑭 : : the ball is blue or green 𝑄 𝐹 = 1 2 Even Event 𝑮 : : the ball is brown or green 𝑄 𝐹 = 2 3 The ball is picked up and we are told that ! Event 𝐹 has occurred. Outco come 𝒀 ! " 𝑄 𝑌 = blue ball = "# = • Event Even 𝑮 gi given en 𝑭 : the ball is brown or green kn known $ tha that the the ball is blue or green # " 𝑄 𝑌 = green ball = "# = • % 𝑄 𝐹|𝐺 = ⋯ % " 𝑄 𝑌 = brown ball = "# = • #

Event Even 𝑭 : : the ball is blue or green 𝑄 𝐹 = 1 2 Even Event 𝑮 : : the ball is brown or green 𝑄 𝐹 = 2 3 The ball is picked up and we are told that ! Event 𝐹 has occurred. Outco come 𝒀 ! " 𝑄 𝑌 = blue ball = "# = • Even Event 𝑮 gi given en 𝑭 : : the ball is brown or green $ kn known that the ball is blue or green # " 𝑄 𝑌 = green ball = "# = • % 𝑄 𝐹|𝐺 = 2 6 = 1 % " 𝑄 𝑌 = brown ball = "# = • 3 #

Discrete Conditional Probability § Suppose we assign a distribution function to a sample space and then learn that an event 𝐹 has occurred. § We call the new probability for an event 𝐺 the co conditional al pr probab ability of 𝐺 gi given en 𝐹 and denote it by 𝑄(𝐺|𝐹) . there is more in informatio ion about the event than was ! available at the beginning of the experiment. How to compute 𝑄(𝐺|𝐹) ? ?

Discrete Conditional Probability § Let Ω = 𝜕 ! , 𝜕 " , ⋯ , 𝜕 # be the original sample space with distribution function 𝑛(𝜕 $ ) assigned. § Given that the event 𝐹 has occurred ( 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 = *(+ ! ) '(&) .

Discrete Conditional Probability § Let Ω = 𝜕 ! , 𝜕 " , ⋯ , 𝜕 # be the original sample space with distribution function 𝑛(𝜕 $ ) assigned. § Given that the event 𝐹 has occurred ( 𝑄(𝐹) > 0 ), the conditional probability 𝑛 𝜕 % 𝐹 = ⋯ 𝜕 ! in 𝐹 𝑛 𝜕 # 𝐹 = 𝑛(𝜕 # ) 𝑄(𝐹) 𝜕 ! not in 𝐹 𝑛 𝜕 # 𝐹 = ⋯

A dice Ex Exper erimen ent Rolling a die once. Sample Space ce 𝛁 Ω = 1, 2, 3, 4, 5, 6 Distribution Funct ction 𝒏(𝝏 𝒌 ) 𝑛 1 = 𝑛 2 = 𝑛 3 = 𝑛 4 = 𝑛 5 = 𝑛 6 = 1 6 Event Even 𝑭 : : 𝒀 > 𝟓 𝑄 𝐹 = 2 6 = 1 3

A dice Even Event 𝑭 : : 𝒀 > 𝟓 𝑄 𝐹 = 2 6 = 1 3 𝑛 𝜕 ! 𝐹 = 𝑛(𝜕 ! ) ? 𝑄(𝐹) Sample Space ce 𝛁 Ω = 1, 2, 3, 4, 5, 6 Distribution Funct ction 𝒏(𝝏 𝒌 ) 𝑛 1 = 𝑛 2 = 𝑛 3 = 𝑛 4 = 𝑛 5 = 𝑛 6 = 1 6

A dice Distribution Funct ction 𝒏(𝝏 𝒌 ) Sample Sa Su Sub-Space ce • 1, 2, 3, 4 𝑛(𝜕 & ) = 1 𝑛 𝜕 ! 𝐹 = 𝑛(𝜕 ! ) ? 6 • 5, 6 𝑄(𝐹) Event Even 𝑭 : : 𝒀 > 𝟓 𝑄 𝐹 = 2 6 = 1 3 Co Conditional Proba babi bility • 𝑛 1 𝐹 = 𝑛 2 𝐹 = 𝑛 3 𝐹 = 𝑛 4 𝐹 = 0 • 𝑛 5 𝐹 = 𝑛 6 𝐹 = "/& "/! = " '

Conditional Probability Formula 𝜕 # in 𝐹 𝜕 # not in 𝐹 𝑛 𝜕 # 𝐹 = 𝑛(𝜕 # ) 𝑛 𝜕 # 𝐹 = 0 𝑄(𝐹) The condition probability of a general event 𝐺 given that 𝐹 occurs, = +(- ! ) /(*) = /((∩*) /(*) . 𝑄 𝐺 𝐹 = ∑ (∩* 𝑛 𝜕 # 𝐹 = ∑ (∩* 𝑄 𝐺 𝐹 = 1(2∩4) 1(4) . =

A dice Even Event 𝑭 : : 𝒀 > 𝟓 Even Event 𝑮 : : 𝒀 is is even 𝑄 𝐺 𝐹 = 1(2∩4) 1(4) . = 𝑄 𝐹 = 2 6 = 1 𝑄 𝐺 = 3 6 = 1 3 2 Even Event 𝑭 ∩ 𝑮 : : 𝒀 = 𝟕 𝑄 𝐹 ∩ 𝐺 = 1 6 Conditional Co Proba babi bility 𝑄 𝐺 𝐹 = 1/6 1/3 = 1 2

A dice Event Even 𝑭 : : 𝒀 > 𝟓 Event Even 𝑮 : : 𝒀 is is a sq square number 𝑄 𝐺 𝐹 = 1(2∩4) 1(4) . = 𝑄 𝐹 = 2 6 = 1 𝑄 𝐺 = 2 6 = 1 3 3 Even Event 𝑭 ∩ 𝑮 : : ∅ 𝑄 𝐹 ∩ 𝐺 = 0 Conditional Co Proba babi bility 𝑄 𝐺 𝐹 = 0 0 2 = 1/3 = 0

Boy or Girl § Suppose you're on a game show, and What is the probability that a you're given the choice of three doors: family of two children has behind one door is a car; behind the others, goats. two boys given that it has at least • one boy? two boys given that the first child • is a boy? one girl given that it has at least • one boy? one girl given that the first child • is a boy?

Boy or Girl § Suppose you're on a game show, and two boys given that it has at least one boy? • you're given the choice of three doors: two boys given that the first child is a boy? • behind one door is a car; behind the one girl given that it has at least one boy? • others, goats. 𝑄 𝐺 𝐹 = 1(2∩4) 1(4) . one girl given that the first child is a boy? • = 𝐺 𝑭 two boys at least one boy one girl the first child is a boy

Boy or Girl Exper Ex erimen ent Sample Space ce 𝛁 A family of two children. Ω = 𝐶𝐶, 𝐶𝐻, 𝐻𝐶, 𝐻𝐻 Distribution Funct ction 𝒏(𝝏 𝒌 ) 𝑛 𝐶𝐶 = 𝑛 𝐶𝐻 = 𝑛 𝐻𝐶 = 𝑛 𝐻𝐻 = 1 1 4 Event: Even at leas east one boy Event: the first ch child is a boy 1 2 𝑄 = ⋯ 𝑄 = ⋯ Event: Even two boys Even Event: one gi girl 1 3 𝑄 = ⋯ 𝑄 = ⋯

Boy or Girl What is the probability that a family of two children has two boys given that it has at least one boy? • 𝑄 𝐺 𝐹 = 1(2∩4) 1(4) . = 1 Even Event 𝑭 : : at least one boy Even Event 𝑮 : two : boys 𝑄 𝐹 = 1 − 1 4 = 3 𝑄 𝐺 = 1 1 2 4 4 Event Even 𝑭 ∩ 𝑮 : : two boys '()∩+) "/! " $ . 𝑄 𝐺 𝐹 = '(+) = $/! = 1 3 𝑄 𝐹 ∩ 𝐺 = 1 4

Boy or Girl What is the probability that a family of two children has two boys given that the first child is a boy? • 𝑄 𝐺 𝐹 = 1(2∩4) 1(4) . = 1 Even Event 𝑭 : : the first is a boy Even Event 𝑮 : two : boys 𝑄 𝐹 = 1 𝑄 𝐺 = 1 1 2 2 4 Event Even 𝑭 ∩ 𝑮 : : two boys '()∩+) "/! " # . 𝑄 𝐺 𝐹 = '(+) = "/# = 1 3 𝑄 𝐹 ∩ 𝐺 = 1 4