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5-3 Addition Rule of Probability Defn: Two events E1 and E2 are mutually exclusive if they cannot occur simultaneous i.e. they have no common outcomes. Ex.: Roll a single die. Let E1 = {event of an even} and E2 = {event of an odd} then E1 = {2,4,6} and E2 = {1,3,5}. E1 and E2 are mutually exclusive-no common events. Ex. Select a single card from a standard deck: Obtain a 7 or a jack. E1 ={ 7?,7?,7?,7? } and E2 = {J? ,J? ,J ? ,J? }. E1 and E2 are mutually exclusive. Obtain a club or a king A = {clubs }, and B ={ kings}. A and B are not mutually exclusive since one King is also a club. Get a face card or get an ace. A = {face card} and B = {ace}. Events A and B are not mutually exclusive. Thrm: (The Addition Rule)- If E1 and E 2 are events then P(E1 or E2 ) = P(E 1 ) + P(E 2 ) - P(E 1 n E 2 ) . E2 E1 Ee Ex: 5-42 page 186 A store has 2 managers, 4 department heads, 16 clerks, 4 stock persons. (n = 26) One person is selected at random. Find the probability that the person is a clerk or a manager. Solution: P(clerk or manager) = 16 + 26 2 - 0 = 18 = .69 26 26 26 HW p 186 #38,39,40,43,48,49,52,57,61 #49 text A Stat class has 18jrs, 10srs, 6srs are Female, 12jrs are Male. Select one student at random, find the probability it is a 1. jr or female 2. sr or female 3. jr or sr. Solve: P(Jr or F) = P(Jr) + P(F) – P(Jr and F) = 18 + 6 – 6 28 28 28 = 18 28 5-4 The Multiplication Rule Defn: Events A and B are independent if A’s occurrence does not affect B’s occurance. Otherwise they are dependent events. Multiplication Rule 1: If A and B are independent events then P(A and B) = P(A nB) = P(A)*P(B). Examples. 1. Flip a coin and toss a die. Find P(Head and a 4). P(Head and a 4 ) = P(H) * P(4) , since flipping the coin and tossing a die are independent events. 2. A card is drawn from a standard deck and replaced; draw a second and replace it. Find the probability of drawing a Queen and an Ace. P(Q and A) = P(Q) * P(A ) = 4 * 4 52 52 3. An urn contains 3 Red, 2 Blue, and 5 White balls. If two are drawn with replacement then: P(B and B) = P(two Blues) = 2 * 2 10 * 10 P(B and W) = 2 * 5 10 * 10 P(R and B) = 3* 2 10 * 10 Multiplication Rule 2 Defn: The conditional probability of event B given event A is the probability that B occurs after event A has already occurred. (Notation P(B |A)) And the conditional probability of B given A is P(B|A) = P(A and B) / P(A) Ex. Draw 2 cards from a standard deck without replacement (dependent events) Find the probability of drawing an Ace and then a King. P(A and K) = P(A) * P(K |A) = 4 * 4 = 4 52 51 663 Note: P(K ?A) = probability of drawing a King (K) given that an Ace(A) has already been drawn. Ex. Given 25 microwave ovens, 2 are known to be defective. Select 2 at random with-out replacement. Find the probability both are defective. Solution: Let D1 = event of a defective on the first draw. D2 = event of defective on 2nd draw. P(D1 and D 2 ) = P(D1 ) * P(D1 /D2 ) = 2 * 1 = 1 25 24 300 Note P(D1 /D2 ) = probability of a defective on the second draw given that the first draw was a defective. Ex. Two cards or drawn without replacement. Find the following probabilities: P(JJ) = P(J)*P(J ?J) = 4 * 3 52 51 P(CS) = P(S)*P(C ?S) = 13 * 13 52 51 P(CC) = P(C)*P(C ?C) = 13 * 12 52 51 J = Jack, C = Club, S = Spade 5-4 More Examples Ex. 5-35 p 195 Draw four (4) cards from a standard deck replacing each card after its drawn. A win secures if at least one ace is drawn. Find the probability of drawing a winning hand. Sol: Let E = {event at least one Ace is drawn} then E = {event no ace is drawn}. So P (E) = 48 * 48 * 48 * 48 = 20,736 52 52 52 52 28,561 So P(E) = 1 – P(E) = 1 – 20,736 = 0.27 28,561 Ex. 5-34 p194 Survey: Should women participate in combat? Select one respondent from the 100. Yes No Total Male 32 18 50 Female 8 42 50 Total 40 60 100 Find the probability the (a) respondent answered Yes given the respondent was female. (b) respondent was male given the respondent answered no. Solve. P(Y ?F) = P(Y and F) = 8/100 = 8 P(F) 50/100 50 P(M ?N) = P(M and N) = 18/100 = 18 P(N) 60/100 60 End

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