251solnH2 2/13/08 (Open this document in 'Page Layout' view!) 1 H. Introduction to Probability 1. Experiments and Probability Text problems 4.1, 4.2. 2. The Venn Diagram and the Addition Rule. Downing & Clark, pg. 96 (pg 85 in 3 rd ed) Basics 1, Application 2, 13. H1 (H0A), Text problems 4.3, 4.4, 4.10, 4.11, 4.8!, 4.9!.[4.3, 4.4, 4.8*, 4.9*]., (4.3,4.4, 4.8*, 4.9*)., H4, H5 (H1, H2). 3. Conditional and Joint Probability, Bayes’ Rule. Text 4.16a-c, 4.18 [4.14a-c, 4.16] (4.13a-c, 4.15). H2, H3 (H0B, H0C), D&C pg 113 (pg 103 in 3 rd ed) 14 (Note error in text - 5/6 of the people in the city support Jones, 5/9 of the people in the country support Jones), 15, 16. H6 (H3). 4. Statistical Independence. Text 4.16d, 4.22, 4.21!, 4.24, 4.30, 4.31, 4.33 [4.14d*, 4.19*, 4.20*, 4.22*, 4.28, 4.29, 4.31, 4.68*] (4.18*, 4.19*, 4.21*, 4.26, 4.27, 4.29). H8, H9 (H5, H6). pg. 97( pg. 85 in 3 rd ed) Applications 4, 5, 8, 9, 10, 11, 44. H7(H4). 5. Review. Section 3 is in this document. ----------------------------------------------------------------------------------------------------. Exercise 4.16 [4.14a-c in 9th] (4.13 a-c in 8th edition): The text gives the following contingency table B B A 10 20 . It asks for the following probabilities: (a) P A B ; (b) P A B and (c) P A B . A 20 40 B A 10 Solution: Add the rows and columns to get A 20 30 B B 10 20 A 2090 40 90 A 90 90 30 90 60 90 B 20 30 . To get the probabilities, divide by 90. 40 60 60 90 B B A .1111 .2222 .3333 90 or . Note that this table consists of 60 A .2222 .4444 .6667 90 .3333 .6667 1.0000 1 30 B B A P A B P A B P A A P A B P A B P A P B PB 1 According to the Instructor’s Solutions Manual (Edited) (a) P(A |B) = 10/30 = 1/3 = 0.3333 (b) P(A | B ) = 20/60 = 1/3 = 0.3333 (c) P( A | B ) = 40/60 = 2/3 = 0.6667 But if we use the multiplication rule: 10 P A B 1 (a) PA B 90 30 P B 3 90 20 P A B 1 (b) P A B 90 60 3 PB 90 40 P AB 2 (c) P A B 90 60 3 PB 90 251solnH2 2/13/08 (Open this document in 'Page Layout' view!) 2 Exercise 4.18 [4.16 in 9th] (4.15 in 8th edition): If P A and B .4 and PB .8 , find P A B . Solution: We can write the Multiplication rule as P A B PB .8 . PA B P A B .4 .5 PB .8 P A B . We know that P A B .4 and P B PROBLEM H2 (Old H0B): Using the same definitions as H1 (H0A) Define Event A as rolling an even number (i.e. the sum of both faces is even) Define Event B as rolling a 5. Define Event C as a 1 on at least one of the dice. a. Find what points are in the events and add the probabilities. P A B , P B A , P A C , P C A , PB C , PC B , P B C , P C B b. Show that the multiplication rule works for A and B . i.e. that P A B P A B PB and that P A B PB AP A c. Do the same for P A C , PB C and P C B . Solution: Define Event A as rolling an even number (i.e. the sum of both faces is even) P A 1 . 2 Define Event B as rolling a 5. PB 4 1 . 36 9 Define Event C as a 1 on at least one of the dice. PC 11 . P B 8 1 PB . 36 9 y 1 2 Diagram for dice problems. x 3 4 5 6 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 4 5 6 4 5 6 7 5 6 7 8 6 7 8 9 7 8 9 10 8 9 10 11 9 10 11 12 a) P A B : This is a conditional probability. This is the probability of rolling an even number, given that we roll a 5. There are 4 points in B . None of them are in A . Thus P A B 0. PB A : This is the probability that we roll a 5, given that we roll an even number. There are 18 points in B . None of then are in A . Thus PB A 0. PA C This is the probability of rolling an even number if we have a 1 on at least one of the dice. There are 11 points in C . These are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (3, 1), (4, 1), (5, 1) and (6, 1). Only 5 of these add to an even number. PA C 5 . 11 PC A : There are 18 points in A . Of these 5, (1, 1), (1, 3), (1, 5), (3, 1) and (5, 1), are in C . PC A 5 18 . 251solnH2 2/13/08 (Open this document in 'Page Layout' view!) 3 PB C : Since C is “a 1 on at least one of the dice” and B is “rolling a 5,” our conditional probability of rolling a 5, assuming that we know there is a one on at least one of the dice, is taken out of the 11 points in C . Only 2 of these points give us a 5, so PB C 2 . 11 PC B : There are only 4 points in B . Of these points only two contain a 1. PC B 2 4 1 2 . P B C : This is the probability of not getting a 5 when you get a 1 on at least one of the dice and is actually the complement of P B C . 9 of the 11 points in C do not give us a 5, so P B C 9 11 . P C B : We are picking points with a one in them from the 32 points in B. We have already found that there are 9 of them. P C B 9 32 . b) We must show that the multiplication rule works for A and B ; i.e. that P A B P A B PB and shown that PA B 0 and PB A 0. So P A B PA B PB 01 0 9 P A B PB AP A 01 0. 2 that P A B P B A P A . We already know that P A B 0 from Problem H0A and we have just and c) We must show that the multiplication rule works for A and C ; i.e. that P A C P C A P A and P A C PA C PC . We already know P A C 5 36 . This is the joint probability of A and C . We have also found the conditional probabilities PA C 5 probabilities P A 1 2 and PC 11 P A C P C AP A 5 18 36 11 and PC A 5 18 and the total . So we can say P A C P A C PC 5 11 1 2 365 . 11 36 365 and Similarly, since we already know P B C 1 , we can use probabilities from part a) to show that 18 2 1 1 PB C PB C PC 2 11 and PB C PC B PB 1 1 . 11 36 36 18 2 9 18 Also, since we already know P C B 1 and PB 4 1 , P B 1 PB 1 1 8 , we can 4 36 9 9 9 9 8 1 use probabilities from part a) to show that P C B P C B P B and 32 9 4 P C B P B C PC 9 11 1 11 36 4 251solnH2 2/13/08 (Open this document in 'Page Layout' view!) 4 PROBLEM H3 (Old H0C):. In H2(H0B), demonstrate Bayes’ rule for P A B and P B A , P A C and PC A , PB C and PC B , P B C and P C B . Solution: P A 1 , PB 1 , PC 11 , P B 8 . 2 9 36 9 PB A 0. PA B 0. Bayes’ Rule says PB A PA B PB P A 9 01 1 0. 2 5 11 PA C PC 11 36 5 . PC A 5 . PA C 5 . PC A 18 18 11 1 P A 2 1 PB C PC 211 11 36 18 1 . PC B 1 . PB C 2 . PC B 2 2 11 1 1 P B 9 9 1 P B C PC 911 11 36 4 9 . PCB 9 . P BC 9 . PCB 32 32 11 8 8 PB 9 9 Downing and Clark, pg. 103, Application 14: In the city 5 6 support Jones, but in the country 5 9 support Jones. Half of the people live in the country and half in the city. What is the probability that a randomly picked Jones supporter lives in the country? Solution: The most important part of this problem is defining events and interpreting probabilities. Use the following definitions for events: J , a person supports Jones; S , a person supports Smith; C , a person lives in the city and Co , a person lives in the country. The corrected problem says P J C 5 6 , PJ Co 5 9 and PC PCo P A B PB PJ Co PCo we need Bayes' Rule. P B A , so PCo J . The hardest part of this is P A PJ finding PJ PJ C PJ Co PJ C PC PJ CoPCo 5 6 1 2 5 9 1 2 512 518 PJ CoPCo 5 9 12 10 2 10 25 15 . So P Co J . 36 36 36 25 1 2 and asks us for P Co J . Since this is a condition-reversing problem, PJ 36 25 5 Downing and Clark, pg. 103, Application 15: If 5% of people with type O blood are left-handed, 10% of people with other blood types are left-handed and 40% have blood type O, what is the probability that a randomly selected left-handed person has blood type O? Solution: Use the following definitions for events: L left-handed, O type O. The problem says PL O .05 , P L O .10 , PO .40 and implies that P O .60 If we do this as a 'box' problem, we can say that, out of 100 people, 40 are type O and 60 not type O. Of the 40 who are type O, 5% or 2 must be left-handed and the remaining 38 are right-handed. Of the 60 who are not type O 10% or 6 are left-handed. Thus out of the 100 people there must be 8 left-handed people of L L 40 O 2 38 whom 2 are type O. so the answer is 2 out of 8 or .25. If we put these in the box we get . 60 O 6 54 More formally, PL PL O PO P L O P O .05 .40 .10 .60 .02 .06 .08 PO L PL O PO P L .05 .40 .02 .25 .08 .08 8 92 100 251solnH2 2/13/08 (Open this document in 'Page Layout' view!) 5 Downing and Clark, pg. 103, Application 16: If 70% of brown-eyed people have brown hair, 20% of green-eyed people have brown hair and 5% of blue-eyed people have brown hair; and, out of the entire population, 75% of people have brown eyes, 5% of people have green eyes and 20% of people have blue eyes, what is the probability that a randomly selected brown-haired person has green eyes? Solution: Use the following definitions for events: BrH Brown-haired, BrE Brown-eyed, GE Greeneyed, BlE Blue-eyed.. The problem says P BrH BrE .70 , P BrH GE .20 , P BrH BlE .05 , PBrE 75 , PGE .05 and PBlE .20 . It is asking for PGE BrH If we do this as a 'box' problem, we can say that, out of 100 people, 75 are brown-eyed, 5 are green-eyed and 20 are blue-eyed. Of the 75 who are brown-eyed, 70% or 52.5 have brown hair. Of the 5 who are greeneyed, 20% or 1 has brown hair. Of the 20 who are blue-eyed, 5% or 1 has brown hair. Thus a total of 52.5 + 1 + 1 = 54.5 have brown hair. Of these BrH BrH 54.5 people only one has green eyes. The BrE 52 .5 22 .5 75 requested probability is one out of 54.5 and 1 GE 1 4 5. .01835 . If we put these in the box to the 54.5 BlE right we get PGE BrH 1 .01835 . 54 .5 1 19 20 54 .5 45 .5 100 More formally, PBrH P BrH BrE PBrE P BrH GE PGE P BrH BlE PBlE .70.75 .20.05 .05.20 PBrH GE PGE .20 .05 .01835 .525 .010 .010 .545 So PGE BrH PBrH .545 PROBLEM H6 (OldH3): Your firm produces 3 different VCRs, The Deluxe, Super Deluxe, and Incredible models. You offer a 1 year warranty. 50% of your sales are Deluxe, 30% are Super Deluxe, and 20% are Incredible. During the warranty period 80% of the Deluxe models fail, 50% of the Super Deluxe models fail, and 30% of the Incredible models fail. Since the Deluxe and Super Deluxe models are not worth repairing, they are replaced when they are returned after failure. To decide on how many repair people to hire, find out what per cent of the returns are incredible. Solution: Let D be the event that a VCR is Deluxe, SD be the event that a VCR is Super Deluxe and I be the event that it is Incredible. Then PD .50,PSD .30 and PI 0.20 . Furthermore, if F represents the event that the VCR fails, P F D .80,P F SD .50 and P F I .30 . From the first set of probabilities we can say that of 100 VCRs, 50 are Deluxe, 30 are Super Deluxe and 20 are Incredible. From the second set of probabilities we can deduce that, of the 50 deluxe VCRs, 80% or 40 fail, of the 30 Super Deluxe VCRs, 50% or 15 fail and, of the 20 Incredible VCRs, 30% or 6 fail. We add these SD Total D I up on the table at left and conclude that there are 40 15 6 61 F a total of 61 failures in the 100 VCRs, of which 6 10 15 14 39 F 6 are incredible so that P I F .0984 . Total 50 30 20 100 61 P F I PI More formally, P I F . But PF PF I PF SD PF D P F PF I PI PF SDPSD PF DPD .30 .20 .50 .30 .80 .50 .61 . So that P I F P F I PI .06 .0984 . P F .61 Parts not copied ©2003 Roger Even Bove