Chapter 15 General Probability Rules 1
Homework p361 2, 3, 4, 10, 11, 12, 30, 36, 38, 40, 42 2
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Objective Students use the general addition and multiplication rules to find probabilities of random events. 4
What if events are not disjoint? When two events A and B are disjoint (mutually exclusive), we can use the addition rule for disjoint events: P(A B) = P(A) + P(B) However, when the events are not disjoint, this earlier addition rule will double count the probability of A and B both occurring. Thus, we need the General Addition Rule. 5
General Addition Rule n Events NOT mutually exclusive n Probability of a 10 OR a club. 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A Note, the 10 is included in both events and would be counted twice. 6
General Addition Rule General Addition Rule: For any two events A and B, P(A B) = P(A) + P(B) P(A B) P(A or B) = P(A) + P(B) P(A and B) 7
NOT Disjoint Events Suppose we have a company with 66 employees. Maintenance - 14, Production - 37, Office - 8, Management - 7 If one employee is chosen at random each month to get a good parking space what is the probability that a maintenance or production worker is chosen? This is very simple IF we assume the positions are mutually exclusive. That is, a maintenance worker is not also management. P(Mt U Mg) = P(Mt) + P(Mg) = 14/66 + 7/66 = 21/66 = 7/22 But what if a maintenance worker is the maintenance supervisor (management)? Then the jobs are not mutually exclusive (disjoint) and: P(Mt U Mg) = P(Mt) + P(Mg) - P(Mt U Mg) = 14/66 + 7/66-1/66 = 20/66 = 10/33 8
It Depends When we looked at contingency tables, we discussed conditional distributions. This leads us to conditional probability. When we want the probability of an event from a conditional distribution, we write P(B A); the probability of B given A. 9
Contingent Probability To find the probability of the event B given the event A, we restrict ourselves to the outcomes in A. We then find the fraction of A outcomes in which B also occurred. P(B \ A) = P(A B) P(A) = P(A and B) P(A) Note: P(A) cannot equal 0, since we know that A has occurred. A1 A2 A3 Total B1 a b c a+b+c B2 d e f d+e+f Total a+d b+e c+f a+b+c+d+e+f P(A1\B1) = P(A2\B2) = P(B1\A1) = P(B2\A3) = a/(a+b+c) e/(d+e+f) a/(a+d) f/(c+f) 10
Example Let us pretend for the moment that 75% of students study at least 3 hours for a statistics test. 50% of students study at least 3 hours AND do all the homework. What is the probability that a student does all the homework given that he studied at least 3 hours for the test. P(Study and HW) =.5, P(Study) =.75 P(Study and HW) = P(Study) P(HW\Study).5 =.75 P(HW\Study) P(HW\Study) = P(Study and HW)/P(Study) =.5 /.75 = 2/3 11
Tables Help Sometimes it helps to put your data into a contingency (table) Mammals Birds Reptiles Amphibians Total US 63 78 14 10 165 Foreign 251 175 64 8 498 Total 314 253 78 18 663 P(Foreign) = 498/663 P(Bird) = 253/663 Foreign and Bird are NOT disjoint P(Foreign Bird) = 175/663 12
Conditional Probability Mammals Birds Reptiles Amphibians Total US 63 78 14 10 165 Foreign 251 175 64 8 498 Total 314 253 78 18 663 P(Foreign) = 498/663 P(Bird) = 253/663 P(Foreign Bird) = 175/663 P(Foreign or Bird) = 498/663 + 253/663-175/663 = 576/663 13
Conditional Probability Mammals Birds Reptiles Amphibians Total US 63 78 14 10 165 Foreign 251 175 64 8 498 Total 314 253 78 18 663 P(Foreign) = 498/663 P(Bird) = 253/663 P(Foreign Bird) = 175/663 P(Foreign or Bird) = 576/663 P(Foreign Bird) = 175/253 P(Bird Foreign) = 175/498 Origin and Animal are NOT independent P(Foreign Bird) = 175/253 P(Foreign) = 498/663 14
Tables Help Mammals Birds Reptiles Amphibians Total US 63 78 14 10 165 Foreign 251 175 64 8 498 Total 314 253 78 18 663 P(Foreign) = 498/663 P(Bird) = 253/663 P(Foreign and Bird) = 175/663 P(Foreign or Bird) = 576/663 P(Foreign Bird) = 175/253 P(Bird Foreign) = 175/498 P(Foreign Bird) = P(Foreign and Bird)/P(Bird) = 175/663 253/663 = 175/253 P(Bird Foreign) = P(Foreign and Bird)/P(Foreign) = 175/663 498/663 = 175/498 15
Or Suppose we have a company with 66 employees. Maintenance - 14, Production - 37, Office - 8, Management - 7 If one employee is chosen at random each month to get a good parking space what is the probability that a maintenance or production worker is chosen? P(M U P) = P(M) + P(P) = 14/66 + 37/66 = 51/66 = 17/22 This, of course, assumes that no employee fills more than one position. The jobs are mutually exclusive (disjoint). 16
Example At our lovely little company, 36 people apply for a job, 20 men and 16 women. Eight of the men and 12 of the women have a Ph.D. If one person is selected at random for an interview, find the probability that the one chosen is a woman or has a Ph.D. P(W or PhD) = P(W) + P(PhD) P(W and PhD) = 16/36 + 20/36-12/36 = 24/36 = 2/3 17
Example 36 people apply for a job, 20 men and 16 women. Eight of the men and 12 of the women have a Ph.D. Find the probability that the one chosen is a woman or has a Ph.D. Ph.D. No Ph.D. Total A contingency table is also useful Men 8 12 20 Women 12 4 16 Total 20 16 36 P(W or PhD) = P(W) + P(PhD) P(W and PhD) = 16/36 + 20/36-12/36 = 24/36 = 2/3 18
Another Example On New Years Eve, the probability that a person driving while intoxicated is 0.32, the probability of a person having a driving accident is 0.09, and the probability of a person having a driving accident while intoxicated is 0.06. What is the probability of a person driving while intoxicated or having a driving accident? P(DI or A) = P(DI) + P(A) P(DI and A) = 0.32 + 0.09-0.06 = 0.35 19
The General Multiplication Rule When two events A and B are independent, we can use the multiplication rule for independent events: P(A B) = P(A) x P(B) 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. P(A B) = P(A) x P(B A) P(A and B) = P(A) x P(B A) or P(A B) = P(B) x P(A B) P(A and B) = P(B) x P(A B) 20
Using a Table Mammals Birds Reptiles Amphibians Total US 63 78 14 10 165 Foreign 251 175 64 8 498 Total 314 253 78 18 663 P(Foreign and Bird) = 175/663 P(Foreign) = 498/663 P(Bird Foreign) = 175/498 P(Foreign and Bird) = P(Foreign) x P(Bird Foreign) = 498/663 x 175/498 = 175/663 21
Independence Independence of two events means that the outcome of one event does not influence the probability of the other. With our new notation for conditional probabilities, we can now formalize this definition: Events A and B are independent whenever P(B A) = P(B). (Equivalently, events A and B are independent whenever P(A B) = P(A).) In other words, to determine if event A and B are independent, determine if P(A B) = P(A). If equal the events are independent. If not equal the events are not independent. Remember, most of our calculations are empirical and estimations of true population values. Thus do not look for exactly equal values. 22
Independent Disjoint Disjoint events cannot be independent! Since we know that disjoint events have no outcomes in common, knowing that one occurred means the other did not occur. Thus, the probability of the second occurring changed based on the fact that the first occurred. Thus, the two events are, most assuredly, not independent. A common error is to confuse independent with disjoint and use the probability rules incorrectly. 23
Mutually Exclusive Independent Comparison Mutually Exclusive Independent Meaning Influence Two events are mutually exclusive, when their occurrence cannot be simultaneous. Occurrence of one event will result in the non-occurrence of the other. Two events are said to be independent, when the occurrence of one event cannot control the occurrence of other. Occurrence of one event will have no influence on the occurrence of the other. Formulae P(A and B) = 0 P(A and B) = P(A) P(B) Venn Diagram No Overlap Overlap 24
Drawing Without Replacement Sampling without replacement means that once an individual is drawn that data value does not go back into the population. This, then, changes the probability for the next selection. Whenever we sample from a population, we sample without replacement. This would suggest all sampling leads to dependent events. Howsomever, if the population is sufficiently large relative to sample size, the difference in probabilities is negligible and the events can be considered independent. 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. 25
Tree Diagrams A tree diagram helps us think through conditional probabilities by showing sequences of events as paths that look like branches of a tree. A tree diagram can help you find the probabilities (or counts) for both independent and dependent events. Suppose a family has two children born apart. 1st child 2nd child boy p = 1/2 2nd child girl p = 1/2 boy p = 1/2 girl p = 1/2 boy p = 1/2 girl p(2 boys) = 1/2 1/2 = 1/4 p(boy/girl) = 1/2 1/2 = 1/4 p(girl/boy) = 1/2 1/2 = 1/4 p(2 girls) = 1/2 1/2 = 1/4 p = 1/2 26
Dependent Events Now let us get real. http://www.in-gender.com/xyu/odds/gender_odds.aspx 1st child 2nd child p =.51 boy p =.50 boy girl p =.50 p(2 boys) =.51.50 =.2550 p(boy/girl) =.51.50 =.2550 girl p =.49 p =.545 boy girl p =.455 p(girl/boy) =.49.545 =.2670 p(2 girls) =.49.455 =.2230 When drawing a tree diagram, completely fill out the diagram. I will ding you if you do not complete the tree. 27
Tree Diagrams Your book has an example of a tree diagram showing the probabilities of drinking and driving accidents. The events drinking and accident are dependent. All the final outcomes are disjoint and must add up to one. We can add the final probabilities to find probabilities of compound events. 28
Example Suppose we have a sack of 12 marbles, 4 blue, 3 white, and 5 red. A marble is selected, color noted, not returned, and a second selected. The appropriate tree would be as follows: 3/11 B P(B & B) =.3333 x.2727 =.0909 B 3/11 W P(B & W) =.3333 x.2727 =.0909 4/12 5/11 R P(B & R) =.3333 x.4545 =.1515 4/11 B P(W & B) =.25 x.3636 =.0909 3/12 W 2/11 W P(W & W) =.25 x.1818 =.0455 5/11 R P(W & R) =.25 x.4545 =.1136 5/12 4/11 B P(R & B) =.4167 x.3636 =.1515 R 3/11 W P(R & W) =.4167 x.2727 =.1136 4/11 R P(R & R) =.4167 x..3636 =.1515 29
Reversing the Conditioning Reversing the conditioning of two events is rarely intuitive, but it is simple algebra. Suppose we want to know P(A B), and we know only P(A), P(B), and P(B A). We also know P(A B), since P(A B) = P(A) x P(B A) From this information, we can find P(A B): P(B A) = P(A B) P(A B) = P(A B) P(A) P(B) 30
Example P(B A) = P(A B) P(A B) = P(A B) P(A) P(B) Mammals Birds Reptiles Amphibians Total US 63 78 14 10 165 Foreign 251 175 64 8 498 Total 314 253 78 18 663 P(F) = 498/663 P(R) = 78/663 P(R&F) = 64/663 P(R F) = 64/498 64 P(R F) = P(R F) P(R) = 663 498 = 64 663 663 498 = 64 498 663 31
Hoyle Everyone is familiar with a deck of cards, correct? 32
Examples A card is drawn from a deck of 52. What is the probability it is an ace or red? Ace and Red are not disjoint P(A or R) = P(A) + P(R) - P(A and R) = 4/52 + 26/52-2/52 P(A or R) = 28/52 = 7/13.5385 Two cards are drawn from a deck of 52 without replacement. What is the probability they are both aces? The draws are NOT independent P(A1 and A2) = P(A1)P(A2 A1) = (4/52)(3/51) = 12/2652.0045 33
Examples Five cards are drawn from a deck of 52. What is the probability all five cards are hearts? The draws are NOT independent P(5 H) = P(H1)P(H2 H1)P(H3 H1 and H2) P(H4 H1, H2, H3)... P(5H) = (13/52)(12/51)(11/50)(10/49)(9/48) =.0005 I draw one card and look at it. I tell you it is red. What is the probability it is a heart? Conditional probability P(Heart Red) = P(H and R)/P(red) = (13/52)/(26/52) = 1/2 34
Examples Let us suppose we notice who is wearing jeans in class one day. The table at right shows the counts. Jeans Other Total Male 12 5 17 Female 8 11 19 Total 20 16 36 What is the probability that a male wears jeans? The events are NOT independent P(J M) = P(J and M)/P(M) = (12/36)/(17/36) = 12/17; P(J) = 20/36 = 10/18 What is the probability someone wearing jeans is a male? The events are NOT independent P(M J) = P(M and J)/P(J) = (12/36)/(20/36) = 12/20 Are being male and wearing jeans disjoint? Explain. 35
Another Tree There is a 30% chance of rain tomorrow. When it is raining PapaOscar takes the bus to work 50% of the time and mommy drives him 50% of the time. When it is not raining PapaOscar takes the bus 25% of the time and mommy brings him 75% of the time. (mommy ain t going out in the rain) What is the probability that PapaOscar (PO) takes the bus tomorrow? 36
Another Tree p =.5 p(rain&bus) =.3.5 =.15 bus p =.3 rain mom p =.5 p(rain&mom) =.3.5 =.15 no rain p =.25 p(no rain&bus) =.7.25 =.175 p =.7 bus mom p =.75 p(no rain&mom) =.7.75 =.525 p(bus) = p(bus&rain OR bus&no rain) p(bus) = p(bus&rain) + p(bus&no rain) =.15 +.175 =.325 The probability that PO takes the bus is.325. 37
Examples In April 2003, Science magazine reported on a new computer based test for ovarian cancer, clinical proteomics, that examines a blood sample for the presence of certain patterns of proteins. Ovarian cancer, though dangerous, is very rare, afflicting only 1 of every 5000 women. The test is highly sensitive, able to correctly detect the presence of ovarian cancer in 99.97% of women who have the disease. However, it is unlikely to be used as a screening test in the general population because the test gave false positives 5% of the time. Why are false positives such a big problem? Draw a tree diagram and determine the probability that a woman who tests positive using this method actually has ovarian cancer. (Notice the difference in the question.) 38
Example Test Positive.0002 x.9997 =.0002 Test.0002 Ovarian Cancer.9997.0003 P(Cancer Positive) = P(Cancer and Positive) P(Positive) Test.0002 x.0003 =.0000 Negative (.0002)(.9997).9998 Test Positive (.0002)(.9997) + (.9998)(.05) No Cancer.05.9998 x.05 =.05.00398.95.9998 x.95 =.9498 Test Negative The vast majority of positive results are false positives. 39
At Least Often it is easier to find the probability of an event not happening (the complement) than it is in finding the probability the event will happen. Then we can subtract from 1 to find the probability of the event. I am not certain how often I must repeat this until it finally sinks in, but when you see the words at least you should consider using the complement. at least 1 not zero Let me repeat that. at least 1 not zero 40
Averting a Nuclear Meltdown In a nuclear plant there are 3 safety shut-off switches. Each switch has a 5% chance of failure. What is the probability that at least one switch will work? At least one switch failing means 1 could fail, or 2 could fail, or all 3 could fail. The only event that is not included in at least one is none. It is much easier to find the probability that none of the switches will work. The 5% chance of failure = p(bad) =.05 P(bad, bad, and bad) =.05.05.05 =.000125 P(at least one good) = 1 -.000125 =.999875 The probability that at least one switch would function properly, averting a nuclear disaster is.999875. What would one more switch do? 41
Review So let us review what we have learned. P(A or B) = P(A) + P(B) P(A and B) When two events A and B are disjoint (mutually exclusive), then P(A and B) = 0 P(A and B) = P(A) x P(B) When two events A and B are independent, then P(B\A) = P(B) 42