foundations of computer science lecture 15...

Foundations of Computer Science

Lecture 15

ProbabilityComputing Probabilities

Probability and Sets: Probability Space

Uniform Probability Spaces

Infinite Probability Spaces

The probable is what usually happens – Aristotle

Last Time

To count complex objects, construct a sequence of “instructions” that can beused to construct the object uniquely. The number of possible sequences ofinstructions equals the number of possible complex objects.

1 Counting Sequences with and without repetition. Subsets with and without repetition. Sequences with specified numbers of each type of object: anagrams.

2 Inclusion-Exclusion (advanced technique).

3 Pigeonhole principle (simple but IMPORTANT technique).

Creator: Malik Magdon-Ismail Probability: 2 / 14 Today →

Today: Probability

1 Computing probabilities.Outcome tree.

Event of interest.

Examples with dice.

2 Probability and sets.The probability space.

3 Uniform probability spaces.

4 Infinite probability spaces.

Creator: Malik Magdon-Ismail Probability: 3 / 14 Chances of Rain →

The Chance of Rain Tomorrow is 40%

What does the title mean? Either it will rain tomorrow or it won’t.

Creator: Malik Magdon-Ismail Probability: 4 / 14 Toss Two Coins →



The chances are 50% that a fair coin-flip will be H.





Flip 100 times. Approximately 50 will be H ← frequentist view.






1 You toss a fair coin 3 times. How many heads will you get?







2 You keep tossing a fair coin until you get a head. How many tosses will you make?







2 You keep tossing a fair coin until you get a head. How many tosses will you make?

There’s no answer. The outcome is uncertain. Probability is appropriate for such settings.


Toss Two Coins: You Win if the Coins Match (HH or TT)

1 You are analyzing an “experiment” whoseoutcome is uncertain.

Creator: Malik Magdon-Ismail Probability: 5 / 14 Event of Interest →



2 Outcomes. Identify all possible outcomes using

a tree of outcome sequences. H T Coin 1





a tree of outcome sequences. H T

H T H T Coin 2

Coin 1





a tree of outcome sequences. H T

H T H T Coin 2

Coin 1

OutcomeHH HT TH TT





a tree of outcome sequences.

3 Edge probabilities. If one of k edges

(options) from a vertex is chosen randomly then

each edge has edge-probability 1k.

H T

H T H T Coin 2

Coin 1

OutcomeHH HT TH TT

1

2

1

2

1

2

1

2

1

2

1

2





a tree of outcome sequences.

3 Edge probabilities. If one of k edges

(options) from a vertex is chosen randomly then

each edge has edge-probability 1k.

4 Outcome-probability. Multiply

edge-probabilities to get outcome-probabilities.

H T

H T H T Coin 2

Coin 1

OutcomeHH HT TH TT

1

2

1

2

1

2

1

2

1

2

1

2

Probability14

14

14

14


Event of Interest

Toss two coins: you win if the coins match (HH or TT)

Question: When do you win? Event: Subset of outcomes where you win.

H T

H T H T Coin 2

Coin 1

OutcomeHH HT TH TT

1

2

1

2

1

2

1

2

1

2

1

2

Probability14

14

14

14

Creator: Malik Magdon-Ismail Probability: 6 / 14 The Outcome-Tree Method →

Event of Interest



5 Event of interest. Subset of the outcomes

where you win.H T

H T H T Coin 2

Coin 1

OutcomeHH HT TH TT

1

2

1

2

1

2

1

2

1

2

1

2

Probability14

14

14

14

HH

14

TT

14


Event of Interest



5 Event of interest. Subset of the outcomes

where you win.

6 Event-probability. Sum of its

outcome-probabilities.

event-probability =1

4+

1

4=

1

2.

H T

H T H T Coin 2

Coin 1

OutcomeHH HT TH TT

1

2

1

2

1

2

1

2

1

2

1

2

Probability14

14

14

14

HH

14

TT

14

Probability that you win is 1

2, written as P[“YouWin”] = 1

2.

Go and do this experiment at home. Toss two coins 1000 times and see how often you win.


The Outcome-Tree Method

Become familiar with this 6-step process for analyzing a probabilistic experiment.

1 You are analyzing an experiment whose outcome is uncertain.

2 Outcomes. Identify all possible outcomes, the tree of outcome sequences.

3 Edge-Probability. Each edge in the outcome-tree gets a probability.

4 Outcome-Probability. Multiply edge-probabilities to get outcome-probabilities.

5 Event of Interest E. Determine the subset of the outcomes you care about.

6 Event-Probability. The sum of outcome-probabilities in the subset you care about.

P[E ] =∑

outcomes ω ∈ EP (ω).

P[E ] ∼ frequency an outcome you want occurs over many repeated experiments.

Pop Quiz. Roll two dice. Compute P[first roll is less than the second].

Creator: Malik Magdon-Ismail Probability: 7 / 14 Let’s Make a Deal →

Let’s Make a Deal: The Monty Hall Problem

1: Contestant at door 1.2: Prize placed behind random door.

1 2 3

Creator: Malik Magdon-Ismail Probability: 8 / 14 Non-Transitive Dice →


1: Contestant at door 1.2: Prize placed behind random door.3: Monty opens empty door (randomly if there’s an option).

1 2 3

1∅

3




1 2 3

1∅

3

Outcome-tree and edge-probabilities.

1

2

3

1

3

1

3

1

3

Prize




1 2 3

1∅

3


1

2

3

2

3

3

2

1

3

1

3

1

3

1

2

1

2

1

1

Prize Host




1 2 3

1∅

3


1

2

3

2

3

3

2

1

3

1

3

1

3

1

2

1

2

1

1

(1, 2)

(1, 3)

(2, 3)

(3, 2)

Prize Host Outcome




1 2 3

1∅

3


Outcome-probabilities.

1

2

3

2

3

3

2

1

3

1

3

1

3

1

2

1

2

1

1

(1, 2)

(1, 3)

(2, 3)

(3, 2)

Prize Host Outcome

P (1, 2) = 1

6

P (1, 3) = 1

6

P (2, 3) = 1

3

P (3, 2) = 1

3

Probability




1 2 3

1∅

3



Event of interest: “WinBySwitching”.

1

2

3

2

3

3

2

1

3

1

3

1

3

1

2

1

2

1

1

(1, 2)

(1, 3)

(2, 3)

(3, 2)

Prize Host Outcome

P (1, 2) = 1

6

P (1, 3) = 1

6

P (2, 3) = 1

3

P (3, 2) = 1

3

Probability




1 2 3

1∅

3



Event of interest: “WinBySwitching”.

Event probability.

1

2

3

2

3

3

2

1

3

1

3

1

3

1

2

1

2

1

1

(1, 2)

(1, 3)

(2, 3)

(3, 2)

Prize Host Outcome

P (1, 2) = 1

6

P (1, 3) = 1

6

P (2, 3) = 1

3

P (3, 2) = 1

3

Probability

1

3+ 1

3= 2

3= P[“WinBySwitching”]


Non-Transitive Dice

A:

B:

C:

Your friend picks a die; you pick a die.e.g. friend picks die B; you pick A.

Creator: Malik Magdon-Ismail Probability: 9 / 14 Probability and Sets →

Non-Transitive Dice

A:

B:

C:


What is the probability that A beats B?


Non-Transitive Dice

A:

B:

C:



Outcome-tree and outcome-probabilities.

Die A

1

3

1

3

1

3


Non-Transitive Dice

A:

B:

C:




Die A Die B

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3


Non-Transitive Dice

A:

B:

C:




Uniform probabilities.

P ( ) = 19

P ( ) = 19

P ( ) = 19

P ( ) = 19

P ( ) = 19

P ( ) = 19

P ( ) = 19

P ( ) = 19

P ( ) = 19

Die A Die B Probability

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

P ( ) = 19

P ( ) = 19

P ( ) = 19

P ( ) = 19

P ( ) = 19


Non-Transitive Dice

A:

B:

C:





Even of interest: outcomes where you win.

P ( ) = 19

P ( ) = 19

P ( ) = 19

P ( ) = 19

P ( ) = 19

P ( ) = 19

P ( ) = 19

P ( ) = 19

P ( ) = 19


1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

P ( ) = 19

P ( ) = 19

P ( ) = 19

P ( ) = 19

P ( ) = 19


Non-Transitive Dice

A:

B:

C:






Number of outcomes where you win: 5.

Probability you win, P[A beats B] = 5

9.

P ( ) = 19

P ( ) = 19

P ( ) = 19

P ( ) = 19

P ( ) = 19

P ( ) = 19

P ( ) = 19

P ( ) = 19

P ( ) = 19


1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

P ( ) = 19

P ( ) = 19

P ( ) = 19

P ( ) = 19

P ( ) = 19


Non-Transitive Dice

A:

B:

C:






Number of outcomes where you win: 5.

Probability you win, P[A beats B] = 5

9.

Conclusion: Die A beats Die B.

Pop Quiz. Compute P[B beats C] and P[C beats A].Hence show A beats B, B beats C and C beats A.

P ( ) = 19

P ( ) = 19

P ( ) = 19

P ( ) = 19

P ( ) = 19

P ( ) = 19

P ( ) = 19

P ( ) = 19

P ( ) = 19


1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

P ( ) = 19

P ( ) = 19

P ( ) = 19

P ( ) = 19

P ( ) = 19


Probability and Sets: The Probability Space

1 Sample Space Ω = ω1, ω2, . . ., set of possible outcomes.2 Probability Function P (·). Non-negative function P (ω), normalized to 1:

0 ≤ P (ω) ≤ 1 and∑

ω∈ΩP (ω) = 1.

Die A

versus B

Ω , , , , , , , ,

P (ω) 1

9

1

9

1

9

1

9

1

9

1

9

1

9

1

9

1

9

Creator: Malik Magdon-Ismail Probability: 10 / 14 Uniform Probability Space →



0 ≤ P (ω) ≤ 1 and∑

ω∈ΩP (ω) = 1.

Die A

versus B

Ω , , , , , , , ,

P (ω) 1

9

1

9

1

9

1

9

1

9

1

9

1

9

1

9

1

9

Events E ⊆ Ω are subsets. Event probability P[E ] is the sum of outcome-probabilities.

“A > B” E1 = , , , ,

“Sum > 8” E2 = , , , ,

“B < 9” E3 = , , , , ,




0 ≤ P (ω) ≤ 1 and∑

ω∈ΩP (ω) = 1.

Die A

versus B

Ω , , , , , , , ,

P (ω) 1

9

1

9

1

9

1

9

1

9

1

9

1

9

1

9

1

9


“A > B” E1 = , , , ,

“Sum > 8” E2 = , , , ,

“B < 9” E3 = , , , , ,

Combining events using logical connectors corresponds to set operations:




0 ≤ P (ω) ≤ 1 and∑

ω∈ΩP (ω) = 1.

Die A

versus B

Ω , , , , , , , ,

P (ω) 1

9

1

9

1

9

1

9

1

9

1

9

1

9

1

9

1

9


“A > B” E1 = , , , ,

“Sum > 8” E2 = , , , ,

“B < 9” E3 = , , , , ,

Combining events using logical connectors corresponds to set operations:“A > B” ∨ “Sum > 8” E1 ∪ E2 = , , , , , , ,

“A > B” ∧ “Sum > 8” E1 ∩ E2 = ,

¬(“A > B”) E1 = , , ,

“A > B”→ “B < 9” E1 ⊆ E3

Important: Exercise 15.9. Sum rule, complement, inclusion-exclusion, union, implication and intersection bounds.


Uniform Probability Space : Probability ∼ Size

P (ω) =1

|Ω|

Creator: Malik Magdon-Ismail Probability: 11 / 14 Poker Probabilities →


P (ω) =1

|Ω|P [E ] =

|E|

|Ω|=

number of outcomes in E

number of possible outcomes in Ω.



P (ω) =1

|Ω|P [E ] =

|E|

|Ω|=



Toss a coin 3 times:

H T H T H T H T

H T H T

H T

Toss 3

Toss 2

Toss 1

Outcome

Probability

HHH

18

HHT

18

HTH

18

HTT

18

THH

18

THT

18

TTH

18

TTT

18

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2



P (ω) =1

|Ω|P [E ] =

|E|

|Ω|=




H T H T H T H T

H T H T

H T

Toss 3

Toss 2

Toss 1

Outcome

Probability

HHH

18

HHT

18

HTH

18

HTT

18

THH

18

THT

18

TTH

18

TTT

18

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

P[“2 heads”] =number of sequences with 2 heads

number of possible sequences in Ω=

3

2

×1

8=

3

8.



P (ω) =1

|Ω|P [E ] =

|E|

|Ω|=




H T H T H T H T

H T H T

H T

Toss 3

Toss 2

Toss 1

Outcome

Probability

HHH

18

HHT

18

HTH

18

HTT

18

THH

18

THT

18

TTH

18

TTT

18

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

P[“2 heads”] =number of sequences with 2 heads

number of possible sequences in Ω=

3

2

×1

8=

3

8.

Practice: Exercise 15.10.

1 You roll a pair of regular dice. What is the probability that the sum is 9?

2 You toss a fair coin ten times. What is the probability that you obtain 4 heads?

3 You roll die A ten times. Compute probabilities for: 4 sevens? 4 sevens and 3 sixes? 4 sevens or 3 sixes?


Poker: Probabilities of Full House and Flush

52 card deck has 4 suits (♠, ♥, ♦, ♣) and 13 ranks in a suit (A,K,Q,J,T,9,8,7,6,5,4,3,2).

Randomly deal 5-cards: each set of 5 cards is equally likely → uniform probability space.

number of possible outcomes =

52

5

possible hands.

Creator: Malik Magdon-Ismail Probability: 12 / 14 Infinite Probability Space →





52

5

possible hands.

Full house: 3 cards of one rank and 2 of another. How many full-houses?






52

5

possible hands.

Full house: 3 cards of one rank and 2 of another. How many full-houses?To construct a full house, specify (rank3, suits3, rank2, suits2). Product rule:






52

5

possible hands.


# full houses = 13×

4

3

×12×

4

2






52

5

possible hands.



4

3

×12×

4

2

→ P[“FullHouse”] =13×

(

4

3

)

× 12×(

4

2

)

(

52

5

) ≈ 0.00144;






52

5

possible hands.



4

3

×12×

4

2


(

4

3

)

× 12×(

4

2

)

(

52

5

) ≈ 0.00144;

Flush: 5 cards of same suit. How many flushes?






52

5

possible hands.



4

3

×12×

4

2


(

4

3

)

× 12×(

4

2

)

(

52

5

) ≈ 0.00144;

Flush: 5 cards of same suit. How many flushes?To construct a flush, specify (suit, ranks). Product rule:






52

5

possible hands.



4

3

×12×

4

2


(

4

3

)

× 12×(

4

2

)

(

52

5

) ≈ 0.00144;


# flushes = 4×

13

5






52

5

possible hands.



4

3

×12×

4

2


(

4

3

)

× 12×(

4

2

)

(

52

5

) ≈ 0.00144;


# flushes = 4×

13

5

→ P[“Flush”] =4×

(

13

5

)

(

52

5

) ≈ 0.00198;

Full house is rarer. That’s why full house beats flush.


Toss a Coin Until Heads: Infinite Probability Space

T

H

1

2

1

2

Toss 1

H

12

Creator: Malik Magdon-Ismail Probability: 13 / 14 Game: First Person To Toss H Wins →


T

H

T

H

1

2

1

2

Toss 1

1

2

1

2

Toss 2

H

12

TH

14



T

H

T

H

T

H

1

2

1

2

Toss 1

1

2

1

2

Toss 2

1

2

1

2

Toss 3

H

12

TH

14

TTH

18



T

H

T

H

T

H

T

H

1

2

1

2

Toss 1

1

2

1

2

Toss 2

1

2

1

2

Toss 3

1

2

1

2

Toss 4

H

12

TH

14

TTH

18

TTTH

116



T

H

T

H

T

H

T

H

T

H

1

2

1

2

Toss 1

1

2

1

2

Toss 2

1

2

1

2

Toss 3

1

2

1

2

Toss 4

1

2

1

2

Toss 5

H

12

TH

14

TTH

18

TTTH

116

TTTTH

132



T

H

T

H

T

H

T

H

T

H

T

H

1

2

1

2

Toss 1

1

2

1

2

Toss 2

1

2

1

2

Toss 3

1

2

1

2

Toss 4

1

2

1

2

Toss 5

1

2

1

2

Toss 6

H

12

TH

14

TTH

18

TTTH

116

TTTTH

132

TTTTTH

164

· · ·

· · ·

· · ·· · ·

Outcome

Probability



T

H

T

H

T

H

T

H

T

H

T

H

1

2

1

2

Toss 1

1

2

1

2

Toss 2

1

2

1

2

Toss 3

1

2

1

2

Toss 4

1

2

1

2

Toss 5

1

2

1

2

Toss 6

H

12

TH

14

TTH

18

TTTH

116

TTTTH

132

TTTTTH

164

· · ·

· · ·

· · ·· · ·

Outcome

Probability

Ω H TH T•2H T•3H T•4H T•5H · · · T•iH · · ·

P (ω) 1

2(1

2)2 (1

2)3 (1

2)4 (1

2)5 (1

2)6 · · · (1

2)i+1 · · ·

# Tosses 1 2 3 4 5 6 · · · i + 1 · · ·



T

H

T

H

T

H

T

H

T

H

T

H

1

2

1

2

Toss 1

1

2

1

2

Toss 2

1

2

1

2

Toss 3

1

2

1

2

Toss 4

1

2

1

2

Toss 5

1

2

1

2

Toss 6

H

12

TH

14

TTH

18

TTTH

116

TTTTH

132

TTTTTH

164

· · ·

· · ·

· · ·· · ·

Outcome

Probability


P (ω) 1

2(1

2)2 (1

2)3 (1

2)4 (1

2)5 (1

2)6 · · · (1

2)i+1 · · ·

# Tosses 1 2 3 4 5 6 · · · i + 1 · · ·

Sum of outcome probabilities:

1

2+ (1

2)2 + (1

2)3 + (1

2)4 + · · · =

∞∑

i=1(1

2)i =

1

2

1− 1

2

= 1.


Game: First Person To Toss H Wins. Always Go First

T

H

T

H

T

H

T

H

T

H

T

H

1

2

1

2

Toss 1

1

2

1

2

Toss 2

1

2

1

2

Toss 3

1

2

1

2

Toss 4

1

2

1

2

Toss 5

1

2

1

2

Toss 6

H

12

TH

14

TTH

18

TTTH

116

TTTTH

132

TTTTTH

164

· · ·

· · ·

· · ·· · ·

Outcome

Probability


P (ω) 1

2(1

2)2 (1

2)3 (1

2)4 (1

2)5 (1

2)6 · · · (1

2)i+1 · · ·

Creator: Malik Magdon-Ismail Probability: 14 / 14


T

H

T

H

T

H

T

H

T

H

T

H

1

2

1

2

Toss 1

1

2

1

2

Toss 2

1

2

1

2

Toss 3

1

2

1

2

Toss 4

1

2

1

2

Toss 5

1

2

1

2

Toss 6

H

12

TH

14

TTH

18

TTTH

116

TTTTH

132

TTTTTH

164

· · ·

· · ·

· · ·· · ·

Outcome

Probability


P (ω) 1

2(1

2)2 (1

2)3 (1

2)4 (1

2)5 (1

2)6 · · · (1

2)i+1 · · ·

The event “YouWin” is E = H, T•2H, T•4H, T•6H, . . ..



T

H

T

H

T

H

T

H

T

H

T

H

1

2

1

2

Toss 1

1

2

1

2

Toss 2

1

2

1

2

Toss 3

1

2

1

2

Toss 4

1

2

1

2

Toss 5

1

2

1

2

Toss 6

H

12

TH

14

TTH

18

TTTH

116

TTTTH

132

TTTTTH

164

· · ·

· · ·

· · ·· · ·

Outcome

Probability


P (ω) 1

2(1

2)2 (1

2)3 (1

2)4 (1

2)5 (1

2)6 · · · (1

2)i+1 · · ·

The event “YouWin” is E = H, T•2H, T•4H, T•6H, . . ..

P[“YouWin”] = 1

2+ (1

2)3 + (1

2)5 + (1

2)7 + · · · = 1

2

∞∑

i=0(1

4)i =

1

2

1− 1

4

= 2

3.

Your odds improve by a factor of 2 if you go first (vs. second).


Today: Conditional Probability

1 New information changes a probability.

2 Definition of conditional probability from regular probability.

3 Conditional probability trapsSampling bias.

Transposed conditional.

4 Law of total probability.Probabilistic case-by-case analysis.

Creator: Malik Magdon-Ismail Conditional Probability: 3 / 16 Flu Season →

Flu Season

1 Chances a random person has the flu is about 0.01 (or 1%) (prior probability).

Probability of flu : P[flu] ≈ 0.01.

Creator: Malik Magdon-Ismail Conditional Probability: 4 / 16 CS, MATH and Dual CS-MATH Majors →

Flu Season



2 You have a slight fever – new information. Chances of flu “increase”.

Probability of flu given fever : P[flu | fever] ≈ 0.4.

New information changes the prior probability to the posterior probability. Translate posterior as “After you get the new information.”

P[A | B] is the (updated) conditional probability of A, given the new information B.


Flu Season



2 You have a slight fever – new information. Chances of flu “increase”.

Probability of flu given fever : P[flu | fever] ≈ 0.4.

New information changes the prior probability to the posterior probability. Translate posterior as “After you get the new information.”

P[A | B] is the (updated) conditional probability of A, given the new information B.

3 Roommie has flu (more new information). Flu for sure, take counter-measures.

Probability of flu given fever and roommie flu : P[flu | fever and roommie flu] ≈ 1.


CS, MATH and Dual CS-MATH Majors

5,000 students: 1,000 CS; 100 MATH; 80 dual MATH-CS.

ALL

CS

MATH

Creator: Malik Magdon-Ismail Conditional Probability: 5 / 16 Conditional Probability P[A | B] →



Pick a random student:

P[CS] = 10005000

= 0.2;

P[MATH] = 1005000

= 0.02;

P[CS and MATH] = 805000

= 0.016.

ALL

CS

MATH





P[CS] = 10005000

= 0.2;

P[MATH] = 1005000

= 0.02;


= 0.016.

New information: student is MATH. What is P[CS |MATH]?Effectively picking a random student from MATH.

ALL

CS

MATH





P[CS] = 10005000

= 0.2;

P[MATH] = 1005000

= 0.02;


= 0.016.


New probability of CS ∼ striped area |CS ∩ MATH|.

ALL

CS

MATH


Conditional Probability P[A | B]

P[A | B] = frequency of outcomes known to be in B that are also in A.

Creator: Malik Magdon-Ismail Conditional Probability: 6 / 16 Chances of Rain →



nB ooutcomes in event B when you repeat an experiment n times.

P[B] =nB

n.





P[B] =nB

n.

Of the nB outcomes in B, the number also in A is nA∩B,

P[A ∩ B] =nA∩B

n.





P[B] =nB

n.


P[A ∩ B] =nA∩B

n.

The frequency of outcomes in A among those outcomes in B is nA∩B/nB,

P[A | B] =nA∩B

nB

=nA∩B

n×

n

nB

=P[A ∩ B]

P[B].





P[B] =nB

n.


P[A ∩ B] =nA∩B

n.

The frequency of outcomes in A among those outcomes in B is nA∩B/nB,

P[A | B] =nA∩B

nB

=nA∩B

n×

n

nB

=P[A ∩ B]

P[B].

P[A | B] =nA∩B

nB

=P[A ∩ B]

P[B]=

P[A and B]

P[B]


Chances of Rain Given Clouds

It is cloudy one in five days, P[Clouds] = 15. It rains one in seven days, P[Rain] = 1

7.

Creator: Malik Magdon-Ismail Conditional Probability: 7 / 16 Conditioning with Dice →



7.

What are the chances of rain on a cloudy day?

P[Rain | Clouds] =P[Rain ∩ Clouds]

P[Clouds].




7.



P[Clouds].

Rainy Days ⊆ Cloudy Days → P[Rain ∩ Clouds] = P[Rain].

All Days

Cloudy

Rainy




7.



P[Clouds].

Rainy Days ⊆ Cloudy Days → P[Rain ∩ Clouds] = P[Rain].

P[Rain | Clouds] =P[Rain]

P[Clouds]=

1715

=5

7.

All Days

Cloudy

Rainy

5-times more likely to rain on a cloudy day than on a random day.

Crucial first step: identify the conditional probability. What is the “new information”?


P[Sum of 2 Dice is 10 | Both are Odd]

Two dice have both rolled odd. What are the chances the sum is 10?

P[Sum is 10 | Both are Odd] =P[(Sum is 10) and (Both are Odd)]

P[Both are Odd]

Creator: Malik Magdon-Ismail Conditional Probability: 8 / 16 Computing a Conditional Probability →




P[Both are Odd]

Probability Space

Die 1 Value

Die

2V

alue





P[Both are Odd]

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

Probability Space

Die 1 Value

Die

2V

alue





P[Both are Odd]

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

Probability Space

Die 1 Value

Die

2V

alue

1 P[Sum is 10] =3

36=

1

12.





P[Both are Odd]

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

Probability Space

Die 1 Value

Die

2V

alue

1 P[Sum is 10] =3

36=

1

12.

2 P[Both are Odd] =9

36=

1

4.





P[Both are Odd]

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

Probability Space

Die 1 Value

Die

2V

alue

1 P[Sum is 10] =3

36=

1

12.


36=

1

4.

3 P[(Sum is 10) and (Both are Odd)] =1

36.





P[Both are Odd]

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

136

Probability Space

Die 1 Value

Die

2V

alue

1 P[Sum is 10] =3

36=

1

12.


36=

1

4.

3 P[(Sum is 10) and (Both are Odd)] =1

36.

4 P[Sum is 10 | Both are Odd] =1

36÷

1

4=

1

9.

Pop Quiz. Compute P[Both are Odd | Sum is 10]. Compare with P[Sum is 10 | Both are Odd].


Computing a Conditional Probability

1: Identify that you need a conditional probability P[A | B].

2: Determine the probability space (Ω, P (·)) using the outcome-tree method.

3: Identify the events A and B appearing in P[A | B] as subsets of Ω.

4: Compute P[A ∩ B] and P[B].

5: Compute P[A | B] =P[A ∩ B]

P[B].

Creator: Malik Magdon-Ismail Conditional Probability: 9 / 16 Monty Prefers Door 3 →

foundations of computer science lecture 15...

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