[PPT] - P( ) 1 conditional probability where P(F) > 0 Conditional PowerPoint Presentation

SLIDE 1 1

Conditional Probability

P( )

SLIDE 2 2

conditional probability

where P(F) > 0

Conditional probability of E given F: probability that E occurs given that F has occurred. “Conditioning on F” Written as P(E|F) Means “P(E, given F observed)” Sample space S reduced to those elements consistent with F (i.e. S ∩ F) Event space E reduced to those elements consistent with F (i.e. E ∩ F)

With equally likely outcomes,

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SLIDE 3

Examples 2 random cards are selected from a deck of cards:

What is the probability that both cards are aces given

that one of the cards is the ace of spades?

What is the probability that both cards are aces given

that at least one of the cards is an ace?

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Gambler’s fallacy Flip a fair coin 51 times A = “first 50 flips are heads” B = “the 51st flip is heads” Pr (B | A) = ?

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conditional probability: the chain rule

General defn: where P(F) > 0 Implies: P(E and F) = P(E|F) P(F) (“the chain rule”) General definition of Chain Rule:

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Chain rule example Alice and Bob play a game as follows: A die is thrown, and each time it is thrown, regardless of the history, it is equally likely to show any of the six numbers. If it shows 5, Alice wins. If it shows 1, 2 or 6, Bob wins. Otherwise, they play a second round and so on. What is P(Alice wins on nth round)?

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SLIDE 7 7

law of total probability

E and F are events in the sample space S

E = EF ∪ EFc

EF ∩ EFc = ∅ ⇒ P(E) = P(EF) + P(EFc) S E F

SLIDE 8 8

law of total probability P(E) = P(E and F) + P(E and Fc) = P(E|F) P(F) + P(E|Fc) P(Fc) = P(E|F) P(F) + P(E|Fc) (1-P(F)) More generally, if F1, F2, ..., Fn partition S (mutually exclusive, ∪i Fi = S, P(Fi)>0), then P(E) = ∑i P(E and Fi) =∑i P(E|Fi) P(Fi)

(Analogous to reasoning by cases; both are very handy.) weighted average, conditioned on event F happening or not. weighted average, conditioned on events Fi happening or not.

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SLIDE 9 9

total probability Sally has 1 elective left to take: either Phys or Chem. She will get A with probability 3/4 in Phys, with prob 3/5 in

Chem. She flips a coin to decide which to take.

What is the probability that she gets an A? P(A) = P(A|Phys)P(Phys) + P(A|Chem)P(Chem) = (3/4)(1/2)+(3/5)(1/2) = 27/40

Note that conditional probability was a means to an end in this example, not the goal itself. One reason conditional probability is important is that this is a common scenario.

SLIDE 10

The Monty Hall Problem

Suppose you’re on a game show, and you’re given the choice of three

doors. Behind one door is a car, behind the other, goats. You pick a

door, say number 1, and the host, who knows what’s behind the doors, opens another door, say number 3, which has a goat. He says to you, “Do you want to switch to door number 2?” Is it to your advantage to switch your choice of doors? Assumptions: The car is equally likely to be behind each of the doors. The player is equally likely to pick each of the three doors, regardless of the car’s location After the player picks a door, the host must open a different door with a goat behind it and offer the player the choice of staying with the original door or switching If the host has a choice of which door to open, then he is equally likely to select each of them.

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Switch

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Independence

[ | ]

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Independence of events Intuition: E is independent of F if the chance of E occurring is not affected by whether F occurs. Formally:

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13

Pr(E|F) = Pr(E) Pr(E ∩ F) = Pr(E)Pr(F)

These two definitions are equivalent.

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Independence Draw a card from a shuffled deck of 52 cards. E: card is a spade F: card is an Ace Are E and F independent?

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Independence Toss a coin 3 times. Each of 8 outcomes equally likely. Define A = {at most one T} = {HHH, HHT, HTH, THH} B = {at most 2 Heads}= {HHH}c Are A and B independent?

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Independence as an assumption It is often convenient to assume independence. People often assume it without noticing. Example: A sky diver has two chutes. Let E = {main chute doesn’t open} Pr (E) = 0.02 F = {backup doesn’t open} Pr (F) = 0.1 What is the chance that at least one opens assuming independence?

16

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Independence as an assumption It is often convenient to assume independence. People often assume it without noticing. Example: A sky diver has two chutes. Let E = {main chute doesn’t open} Pr (E) = 0.02 F = {backup doesn’t open} Pr (F) = 0.1 What is the chance that at least one opens assuming independence? Note: Assuming independence doesn’t justify the assumption! Both chutes could fail because of the same rare event, e.g. freezing rain.

17

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Using independence to define a probabilistic model We can define our probability model via independence. Example: suppose a biased coin comes up heads with probability 2/3, independent of other flips. Sample space: sequences of 3 coin tosses. Pr (3 heads)=? Pr (3 tails) = ? Pr (2 heads) = ?

18

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SLIDE 19 19

Bayes Theorem

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SLIDE 20

Is your coin loaded? Your coin is fair (Pr(H) = 0.5) with probability ½ or “unfair” (Pr(H) = 0.6) otherwise. You flip the coin and it comes up Heads. What is the probability that it is fair given that it came up Heads?

20

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Most common form: Expanded form (using law of total probability):

Bayes Theorem

Why it’s important: Reverse conditioning P( model | data ) ~ P( data | model ) Combine new evidence (E) with prior belief (P(F)) Posterior vs prior

Rev. Thomas Bayes c.

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SLIDE 22 22

HIV testing

Suppose an HIV test is 98% effective in detecting HIV, i.e., its “false negative” rate = 2%. Suppose furthermore, the test’s “false positive” rate = 1%. 0.5% of population has HIV Let E = you test positive for HIV Let F = you actually have HIV What is P(F|E) ? HIV+ HIV- Test + 0.98 = P(E|F) 0.01 = P(E|Fc) Test - 0.02 = P(Ec|F) 0.99 = P(Ec|Fc)

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SLIDE 23 23

HIV testing

Suppose an HIV test is 98% effective in detecting HIV, i.e., its “false negative” rate = 2%. Suppose furthermore, the test’s “false positive” rate = 1%. 0.5% of population has HIV Let E = you test positive for HIV Let F = you actually have HIV What is P(F|E) ? Solution:

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Why do you have a fever? Pr (flu | high fever)?

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SLIDE 25 25

SLIDE 26 A lie detector is known to be 80% reliable when the person is guilty and 95% reliable when the person is innocent. If a suspect is chosen from a group of suspects of which only 2% have ever committed a crime, and the test indicates that the person is guilty, what is the probability that he is innocent? 26

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SLIDE 27

A lie detector is known to be 80% reliable when the person is guilty and 95% reliable when the person is innocent. If a suspect is chosen from a group of suspects of which only 2% have ever committed a crime, and the test indicates that the person is guilty, what is the probability that he is innocent?

27

SLIDE 28

Problem There is a population of N people. The number of good guys among these people is i with probability pi Take a sample of n people from the population. Each subset

f n is equally likely. What is the probability that there are

j good guys in the population conditioned on the fact that there are k good guys in the sample.

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SLIDE 30

Problem There is a population of N people. The number of good guys among these people is i with probability pi Take a sample of n people from the population. What is the probability that there are j good guys in the population conditioned on the fact that there are k good guys in the sample.

29

Ei = event that there are i good guys among N Si = event that there are i good guys in sample Pr(Ej|Sk) = Pr(Sk|Ej)Pr(Ej) Pr(Sk) Pr(Sk|Ej) = j

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SLIDE 31

Conditional Probability Satisfies usual axioms of probability Example: Pr( E | F ) = 1- Pr (Ec | F)

31

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SLIDE 32

Conditional Probabilities yield a probability space Suppose that is a probability space. Then is a probability space for with

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Pr(F) > 0 0 ≤ Pr(w|F) ≤ 1 Pr(∪n

i=1Ei|F) = n

X

i=1

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SLIDE 33

summary

33

SLIDE 34 34

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SLIDE 35 35

more HIV testing Let Ec = you test negative for HIV Let F = you actually have HIV What is P(F|Ec) ?

HIV+ HIV- T est + 0.98 = P(E|F) 0.01 = P(E|Fc) T est - 0.02 = P(Ec|F) 0.99 = P(Ec|Fc)