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Rigollet fall 2012 if two events are not disjoint

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Unformatted text preview: neral addition rule ORF 245 Prof. Rigollet Fall 2012 If two events are not disjoint, there is still a rule: P (A B ) = P (A) + P (B ) P (A ⇥ B ) If A and B are disjoint, it is impossible to have an outcome in A B and thus P (A B ) = 0 This rule can be illustrated with the following diagram: Two layers A on the B intersection ORF 245 Prof. Rigollet Fall 2012 Equally likely outcomes We already know that for equally likely outcomes: 1 P(outcome) = number of outcomes Using the addition rule for disjoint events we find that for an event E = { 1 , . . . , k } = { 1 } · · · { k } P (E ) = P ( 1) + ··· + P( k) = kP ( 1 ) number of outcomes in E = total number of outcomes Need to count outcomes (in E and total) ORF 245 Prof. Rigollet Fall 2012 Counting rules Product rule. If the experiment consists of n1 several consecutive parts where the first part has outcomes, the second part has n2 outcomes,... then the total number of outcomes is given by the product n1 · n2 · · · Example. Rolling two dice. What is the probability of getting two numbers less than or equal to 2? •Total number of outcomes: 6*6=36 •Number of outcomes in event: 2*2=4 Answer is 4/36=1/9 Counting rules ORF 245 Prof. Rigollet Fall 2012 The product rule can take you a long way but sometimes the picture is more subtle. Example. Drawing 5 cards from a 52-card deck. How many hands are there? We could use the product rule: 52 ways of choosing the first card, 51 ways of choosing the second card,... 52*51*50*49*48 ORF 245 Prof. Rigollet Fall 2012 Counting rules We have counted the same hand several times. How many? 5 positions for , 4 positions for that’s 5*4*3*2*1. , 3 positions for Finally we obtain that the number of such hands is 52 · 51 · 50 · 49 · 48 5·4·3·2·1 , ... ORF 245 Prof. Rigollet Fall 2012 Counting rules For the general case: define “factorial k” by k ! = k · (k We have 1) · (k 2) · · · 2 · 1 52! 52 · 51 · 50 · 49 · 48 52! 47! = = 5·4·3·2·1 5! (52 5)!5! Combination rule. The number of outcomes (combinations) obtained when selecting k different objects from a set of n objects is given by n k ⇥ n! := (n k )!k ! “n choose k” ORF 245 Prof. Rigollet Fall 2012 Counting rules Exercise. Three balls are selected at random from the jar below. What is the probability of getting exactly one red ball and two green balls? ⇥ 8 8! Total number of outcomes: = = 56 3 5!3! Number of outcomes in the event {1 red, 2 greens}: Choosing the red ball: 2 possibilities ⇥ Choosing the...
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This document was uploaded on 10/14/2013.

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