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Unformatted text preview: CS70: Satish Rao: Administration. 1. Midterm regrades requests due by your section. CS70: Satish Rao: Lecture 20. 1. Product Rule. 2. Product Rule and Probability spaces. 3. Independence. 4. Inclusion/Exclusion. Product Rule What is Pr [ A ∩ B ] ? Pr [ A ∩ B ] = Pr [ A | B ] Pr [ B ] Example: A- red die is 1 Product Rule What is Pr [ A ∩ B ] ? Pr [ A ∩ B ] = Pr [ A | B ] Pr [ B ] Example: A- red die is 1 B- sum of die is 4 Product Rule What is Pr [ A ∩ B ] ? Pr [ A ∩ B ] = Pr [ A | B ] Pr [ B ] Example: A- red die is 1 B- sum of die is 4 Pr [ A ∩ B ] == 1 3 × 3 36 = 1 36 Intersection of Events. In general, for events A 1 ,..., A n , Pr [ A 1 ∩ A 2 ···∩ A n ] = Pr [ A 1 ] Pr [ A 2 | A 1 ] ··· Pr [ A n | A 1 ∩ A 2 ···∩ A n- 1 ] Intersection of Events. In general, for events A 1 ,..., A n , Pr [ A 1 ∩ A 2 ···∩ A n ] = Pr [ A 1 ] Pr [ A 2 | A 1 ] ··· Pr [ A n | A 1 ∩ A 2 ···∩ A n- 1 ] Proof: Induction. Base Case: plug B = A 1 and A = A 2 into Pr [ A | B ] = Pr [ A | B ] Pr [ B ] . Intersection of Events. In general, for events A 1 ,..., A n , Pr [ A 1 ∩ A 2 ···∩ A n ] = Pr [ A 1 ] Pr [ A 2 | A 1 ] ··· Pr [ A n | A 1 ∩ A 2 ···∩ A n- 1 ] Proof: Induction. Base Case: plug B = A 1 and A = A 2 into Pr [ A | B ] = Pr [ A | B ] Pr [ B ] . Induction: Let A = A n , and B = A 1 ∩ A 2 ∩ A 3 ··· A n- 1 . Pr [ B ∩ A ] = Pr [ B ] Pr [ A | B ] Intersection of Events. In general, for events A 1 ,..., A n , Pr [ A 1 ∩ A 2 ···∩ A n ] = Pr [ A 1 ] Pr [ A 2 | A 1 ] ··· Pr [ A n | A 1 ∩ A 2 ···∩ A n- 1 ] Proof: Induction. Base Case: plug B = A 1 and A = A 2 into Pr [ A | B ] = Pr [ A | B ] Pr [ B ] . Induction: Let A = A n , and B = A 1 ∩ A 2 ∩ A 3 ··· A n- 1 . Pr [ B ∩ A ] = Pr [ B ] Pr [ A | B ] Pr [ A 1 ∩··· A n ] = Pr [ A 1 ∩ A n- 1 ] Pr [ A n | A 1 ∩··· A n- 1 ] Intersection of Events. In general, for events A 1 ,..., A n , Pr [ A 1 ∩ A 2 ···∩ A n ] = Pr [ A 1 ] Pr [ A 2 | A 1 ] ··· Pr [ A n | A 1 ∩ A 2 ···∩ A n- 1 ] Proof: Induction. Base Case: plug B = A 1 and A = A 2 into Pr [ A | B ] = Pr [ A | B ] Pr [ B ] . Induction: Let A = A n , and B = A 1 ∩ A 2 ∩ A 3 ··· A n- 1 . Pr [ B ∩ A ] = Pr [ B ] Pr [ A | B ] Pr [ A 1 ∩··· A n ] = Pr [ A 1 ∩ A n- 1 ] Pr [ A n | A 1 ∩··· A n- 1 ] = ( Pr [ A 1 ] Pr [ A 2 | A 1 ] ··· Pr [ A n- 1 | A 1 ··· A n- 2 ]) × Pr [ A n | A 1 ∩ A 2 ···∩ A n- 1 ] Example. Probability of a flush? Experiment: a deal of a poker hand. Example. Probability of a flush? Experiment: a deal of a poker hand. A 1- second card is same suit as first. Example. Probability of a flush? Experiment: a deal of a poker hand....
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