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W HICH OF THE FOLLOW STATISTICS IS NOT TRUE ? Slide More than 50% of American adults are single. 2. For the average individual, the number of years spent unmarried now outweighs the number of years married. 3. Compared with their married counterparts, single people are more likely to eat out and exercise, go to art and music classes, attend public events and lectures, and volunteer. 4. None of the above.

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U PCOMING I N C LASS Wednesday’s Quiz 2, covers HW3 of the material learned in class. (9/11) Sunday HW4 (9/15) Part 2 of the data project due next Monday (9/16)

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C HAPTER 13 Probability General Addition and Multiplication Rules. Non-disjoint and dependent events. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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T HE G ENERAL A DDITION R ULE General Addition Rule: For any two events A and B, P ( A or B ) = P ( A ) + P ( B ) – P ( A and B ) The following Venn diagram shows a situation in which we would use the general addition rule: Slide 1- 4

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H OMEWORK P ROBLEM A check of dorm rooms on a large college campus revealed that 39% had refrigerators 60% had TVs 20% had both a refrigerator and TV

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W HAT ’ S THE PROBABILITY THAT A RANDOMLY SELECTED DORM ROOM HAS A REFRIGERATOR BUT NO TV? Slide

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W HAT ’ S THE PROBABILITY OF HAVING A REFRIGERATOR OR A TV BUT NOT BOTH ? Slide

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W HAT ’ S THE PROBABILITY OF HAVING NEITHER A REFRIGERATOR NOR A TV ? Slide ( ) 4. 1-( )

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T HE G ENERAL A DDITION R ULE General Addition Rule: For any two events A and B, P ( A or B ) = P ( A ) + P ( B ) – P ( A and B ) For disjoint events P(A and B) =0. Slide 1- 9

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M ULTIPLICATION R ULE F OR I NDEPENDENT E VENTS When two events A and B are independent, we can use the multiplication rule for independent events from Chapter 14: P ( A and B ) = P ( A ) x P ( B ) However, when our events are not independent, this earlier multiplication rule does not work. Thus, we need the General Multiplication Rule. Slide 1- 10

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T HE G ENERAL M ULTIPLICATION R ULE We encountered the general multiplication rule in the form of conditional probability. Rearranging the equation in the definition for conditional probability, we get the General Multiplication Rule: For any two events A and B, P ( A and B ) = P ( A ) x P ( B | A ) or P ( A and B ) = P ( B ) x P ( A | B ) Slide 1- 11

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I T D EPENDS … Back in Chapter 3, we looked at contingency tables and talked about conditional distributions. When we want the probability of an event from a conditional distribution, we write P ( B | A ) and pronounce it “the probability of B given A.” A probability that takes into account a given condition is called a conditional probability. Slide 1- 12

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C ONDITIONAL D ISTRIBUTIONS A conditional distribution shows the distribution of one variable for just the individuals who satisfy some condition on another variable. The following is the conditional distribution of ticket Class, conditional on having survived:

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C ONDITIONAL D ISTRIBUTIONS ( CONT.) The following is the conditional distribution of ticket Class, conditional on having perished: Slide 3- 14

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C ONDITIONAL P ROBABILITIES You draw a card at random from a standard deck of 52 cards. 4 suits (hearts, clubs, spades, diamond) 26 red 26 black 12 face cards 4 Aces Slide 1- 15

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W HAT ’ S THE PROBABILITY THAT A CARD IS A HEART, GIVEN THAT IT IS RED ? Slide / / / /26

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W HAT ’ S THE PROBABILITY THAT THE CARD IS RED, GIVEN THAT IT IS A HEART ? Slide / / /13

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W HAT ’ S THE PROBABILITY THAT THE CARD IS A TEN, GIVEN THAT IT IS RED ? Slide / / / /26

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W HAT ’ S THE PROBABILITY THAT THE CARD IS A KING, GIVEN THAT IT IS A FACE CARD ? Slide / / / /12

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T HE G ENERAL M ULTIPLICATION R ULE We encountered the general multiplication rule in the form of conditional probability. Rearranging the equation in the definition for conditional probability, we get the General Multiplication Rule: For any two events A and B, P ( A and B ) = P ( A ) x P ( B | A ) or P ( A and B ) = P ( B ) x P ( A | B ) Slide 1- 20

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D RAWING W ITHOUT R EPLACEMENT Sampling without replacement means that once one object is drawn it doesn’t go back into the pool. We often sample without replacement, which doesn’t matter too much when we are dealing with a large population. However, when drawing from a small population, we need to take note and adjust probabilities accordingly. Drawing without replacement is just another instance of working with conditional probabilities. Slide 1- 21

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Y OU ARE DEALT A HAND OF THREE CARDS, ONE AT A TIME. W HAT ’ S THE PROBABILITY THE FIRST RED CARD YOU GET IS THE 3 RD CARD DEALT ? 1. ½ + ½ + ½ 2. ½ * ½ * ½ 3. 26/52 * 25/51 * 26/ /52 * 26/52 * 26/52 Slide 1- 22

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W HAT IS THE PROBABILITY THAT YOUR CARDS ARE ALL HEARTS ? 1. 13/52 * 12/51 * 11/ /52 * 25/51 * 24/50 3. ¼ * ¼ * ¼ 4. ½ * ½ * ½ Slide 1- 23

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W HAT ’ S THE PROBABILITY THAT YOU GET NO BLACK CARDS ? 1. ¾ * ¾ * ¾ 2. ¼ * ¼ * ¼ 3. 39/52 * 38/51 * 37/ /52 * 25/51 * 24/50 Slide 1- 24

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W HAT ’ S THE PROBABILITY THAT YOU HAVE AT LEAST ONE SPADE ? /52 * 12/51 * 11/ ¼ * ¼ * ¼ ¼ /52 * 38/51 * 37/50 Slide 1- 25

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T REE D IAGRAMS Slide A tree diagram helps us think through conditional probabilities by showing sequences of events as paths that look like branches of a tree. Figure 15.4 is a nice example of a tree diagram and shows how we multiply the probabilities of the branches together:

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H OMEWORK PROBLEM Mary is flying from Boston to Denver with a connection in Chicago. The probability her first flight leaves on time is 0.2. If the flight is on time, the probability that her luggage will make the connecting flight is 0.95, but if the flight is delayed, the probability that the luggage will make it is only Slide 1- 27

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A RE THE FOLLOWING EVENTS INDEPENDENT - FLIGHT LEAVING ON TIME AND THE LUGGAGE MAKING IT TO D ENVER WITH HER ? Slide Yes, the luggage is equally likely to arrive with her whether or not the flight is on time. 2. No, the probability that the luggage arrives with her depends on whether the flight is on time.

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W HAT ’ S THE PROBABILITY THAT HER LUGGAGE ARRIVES IN D ENVER WITH HER ? Slide

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I T D EPENDS … ( CONT.) To find the probability of the event B given the event A, we restrict our attention to the outcomes in A. We then find in what fraction of those outcomes B also occurred. Note: P ( A ) cannot equal 0, since we know that A has occurred. Slide 1- 30

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S OBRIETY CHECKPOINT PROBLEM Police establish a sobriety checkpoint where they detain drivers whom they suspect have been drinking and release those who have not. The police detain 81% of drivers who have been drinking and release 81% of drivers who have not. Assume that 10% of drivers have been drinking. Slide 1- 31

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W HAT ’ S THE PROBABILITY OF ANY GIVEN DRIVER WILL BE DETAINED ? Slide *.81+.9* *.19+.9* * *.81

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W HAT ’ S THE PROBABILITY THAT A DRIVER WHO IS DETAINED HAS ACTUALLY BEEN DRINKING ? / / / / 25.2 Slide 1- 33

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W HAT ’ S THE PROBABILITY THAT A DRIVER WHO WAS RELEASED HAD ACTUALLY BEEN DRINKING ? / / / / 25.2 Slide 1- 34

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O NE MORE T REE P ROBLEM Dan’s Diner employs three dishwashers. Al washes 60% of the dishes and breaks only 2% of those he handles. Betty and Chuck each wash 20% of the dishes, and Betty breaks only 2% of hers, but Chuck breaks 4% of the dishes he washes. You go to Dan’s for supper one night and hear a dish break at the sink. Slide 1- 35

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W HAT ’ S THE PROBABILITY C HUCK BROKE THE DISH ? Slide /( ) /( ) /(.04) /(.96)

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W HAT ’ S THE PROBABILITY THAT A L BROKE THE DISH ? Slide /( ) /( ) /(.04) /(.96)

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U PCOMING I N C LASS Wednesday’s Quiz 2, covers HW3 of the material learned in class. (9/11) Sunday HW4 (9/15) Part 2 of the data project due next Monday (9/16)

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