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Review

Probability Axioms – Non-negativity P(A)≥0

– Additivity P(A U B) =P(A)+ P(B), if A and B are disjoint.

– Normalization P(Ω)=1

Independence of two events A and B– P(A∩B)=P(A)P(B)

– Two coin tosses A={first toss is a head}, B={second toss is a head}

– Disjoint vs independent P(A∩B)=0, if P(A)>0 and P(B)>0, P(A∩B) < P(A)P(B). They are never

independent. A={first toss is a head}, B={first toss is a tail}

P(A)=0.5, P(B)=0.5, P(A∩B)=0

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1.3 Conditional Probability 1.4 Total Probability Theorem and Bayes’

Rule

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Conditional Probability

A way to reason about the outcome given partial information Example1

– To toss a fair coin 100 times, what’s the probability that the first toss was a head?

Fair coin 1/2

– To toss a fair coin 100 times, if 99 tails come up, what’s the probability that the first toss was a head?

Very small?

Example2– A fair coin and an unfair coin (1/4 tail, 3/4 head)

The first toss is fair, if the outcome is a head, use the fair coin for the 2nd toss, if the outcome is a tail, use the unfair coin for the 2nd toss.

What’s the probability that the 2nd toss was a tail? – ½x½ + ½x¼ = 0.375

What’s the probability that the 2nd toss was a tail if we know that the first toss was a tail?

– 1/4

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Conditional Probability

Definition of a conditional probability– The probability of event A given event B (P(B)>0 )

– P(A|B)=P(A) if A and B are

independent

A new probability law (recall the definition of probability laws)–

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Conditional Probability

Examples – Two rolls of a die, what’s the probability that the first roll was a 1?

– Fair dice 1/6

– Two rolls of a die, the sum of the two rolls is 6, what’s the probability that the first roll was a 1?

B: (1,5) (2,4) (3,3) (4,2) (5,1) , A and B: (1,5) P(A|B)= (1/36)/(5/36)=1/5

– Two rolls of a die, the sum of the two rolls is 6, what’s the probability that the first roll was EVEN?

B: (1,5) (2,4) (3,3) (4,2) (5,1) , A and B: (2,4) (4,2) P(A|B)= (2/36)/(5/36)=2/5

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Conditional Probability

The new universe is B P(A1)> P(A2), does it mean that P(A1|B)> P(A2|B)? No!

– An Example: Two rolls of a die B: the sum of the two rolls is 4, (1,3) (2,2) (3,1) A1: the first roll was 1 or 2 A2: the first roll was 3, 4, 5 or 6

– P(A1 )=1/3 P(A2)=2/3 P(B)= 3/36 =1/12

– P(A1 ∩ B) = 2/36 = 1/18 P(A2 ∩ B) = 1/36

– P(A1 | B)= (1/18)/(1/12) =2/3 P(A2 | B)= (1/36)/(1/12) =1/3

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Conditional Probability

The Chain Rule

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Conditional Independence

Conditional independence, A and C are independent conditional on B, P(B)>0

P(A∩C|B)=P(A|B) P(C|B) Example (conditional independence ≠ independence): unfair

coins, coin 1- (0.9 head, 0.1 tail) coin 2- (0.1 head, 0.9 tail), coin 3 is fair.– Toss coin 3 first. If it’s head, toss coin 1 twice. If it’s tail, toss coin 2

twice.

– A= X H X, the event that the 2nd toss is a head

– C= X X H, the event that the 3rd toss is a head

– B= H X X, the event that the first toss is a head

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Total Probability Theorem

A1 , A2, … An be a partition of Ω– Recall the definition of a partition

Total Probability Theorem

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Total Probability Theorem

An example– A fair coin and an unfair coin (1/4 tail, 3/4 head)

The first toss is fair, if the outcome is a head, use the fair coin for the 2nd and 3rd toss, if the outcome is a tail, use the unfair coin.

– B={ the 2nd and 3rd tosses are both tails}

– A1 ={the first toss is an head}, A2 ={the first toss is a tail}. A1 and A2 is a partition of the universe.. P(A1)=P(A2)= 1/2

– P(B|A1 )= 1/4, P(B|A2 )= 1/16

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Bayes’ Rule

A1 , A2, … An be a partition of Ω

Bayes’ Rule

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Bayes’ Rule

An Example

Question: – How likely is there a tumor given that a shade is observed?

– P(A2 |B)

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Bayes’ Rule

Bayes’ Rule from scratch

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Sending a bit through a noisy channel

Sender has a bit b- either 0 or 1 with equal probability to send to the receiver

– p=0.1

– Question1: if the sender sends b once, and the receiver receives 1, what can the receiver say about b?

– Question2: if the sender sends b 3 times, and the receiver receives 1,1,1 what can the receiver say about b?

– Question3: if the sender sends b 3 times, and the receiver receives 1,0,1 what can the receiver say about b?