Test 1 Results You will get back your test 1 papers on Friday. There is a generous nonlinear curve of the scores: if x is your raw score, your grade out of 100 can be computed as 100x x x = 10 44 = 50 11 1 15 2 21 3 26 4 30 5 34 6 37 7 40 8 43 9 45 10 48 11 50 12 52 13 54 14 56 15 58 16 60 17 62 18 64 19 66 20 67 21 69 22 71 23 72 24 74 25 75 26 77 27 78 28 80 29 81 30 83 31 84 32 85 33 87 34 88 35 89 36 90 37 92 38 93 39 94 40 95 41 97 42 98 43 99 44 100 Arthur Berg 2/ 11
Grade Distributions, Old and New Arthur Berg 3/ 11
Unemployment Rate 9.4 10 6 (job-seekers) 154 10 6 (job-seekers + employed) = 6.1% Arthur Berg 4/ 11
Unemployment by Education Arthur Berg 5/ 11
Relative Frequencies of Employment by Education Arthur Berg 6/ 11
Example 3.1 Conditional distribution on Men and conditional distribution on Women: Arthur Berg 7/ 11
Probability of Two Girls 1 What is the probability that a family with two children has two girls? 2 If you know the given family has at least one girl, what is the probability of the family having two girls? Definition If A and B are any two events, then the conditional probability of A given B, denoted by P(A B), is P(A B) = P(AB) P(B) provided that P(B) > 0. P(AB) can then be computed in two ways as: P(AB) = P(A B)P(B) P(AB) = P(B A)P(A) Arthur Berg 8/ 11
Probability of Two Girls 1 What is the probability that a family with two children has two girls? S = {BB, BG, GB, GG} where each is equally likely, so the probability is 1/4. 2 If you know the given family has at least one girl, what is the probability of the family having two girls? Definition If A and B are any two events, then the conditional probability of A given B, denoted by P(A B), is P(A B) = P(AB) P(B) provided that P(B) > 0. P(AB) can then be computed in two ways as: P(AB) = P(A B)P(B) P(AB) = P(B A)P(A) Arthur Berg 8/ 11
Probability of Two Girls 1 What is the probability that a family with two children has two girls? S = {BB, BG, GB, GG} where each is equally likely, so the probability is 1/4. 2 If you know the given family has at least one girl, what is the probability of the family having two girls? Definition If A and B are any two events, then the conditional probability of A given B, denoted by P(A B), is P(A B) = P(AB) P(B) provided that P(B) > 0. P(AB) can then be computed in two ways as: P(AB) = P(A B)P(B) P(AB) = P(B A)P(A) Arthur Berg 8/ 11
Probability of Two Girls 1 What is the probability that a family with two children has two girls? S = {BB, BG, GB, GG} where each is equally likely, so the probability is 1/4. 2 If you know the given family has at least one girl, what is the probability of the family having two girls? S = {BG, GB, GG} where each is equally likely, so the probability is 1/3. Definition If A and B are any two events, then the conditional probability of A given B, denoted by P(A B), is P(A B) = P(AB) P(B) provided that P(B) > 0. P(AB) can then be computed in two ways as: P(AB) = P(A B)P(B) P(AB) = P(B A)P(A) Arthur Berg 8/ 11
Probability of Two Girls 1 What is the probability that a family with two children has two girls? S = {BB, BG, GB, GG} where each is equally likely, so the probability is 1/4. 2 If you know the given family has at least one girl, what is the probability of the family having two girls? S = {BG, GB, GG} where each is equally likely, so the probability is 1/3. Definition If A and B are any two events, then the conditional probability of A given B, denoted by P(A B), is P(A B) = P(AB) P(B) provided that P(B) > 0. P(AB) can then be computed in two ways as: P(AB) = P(A B)P(B) P(AB) = P(B A)P(A) Arthur Berg 8/ 11
Example 3.2 Example (Example 3.2, p. 64) There are four batteries, and one is defective. Two are to be selected at random for use on a particular day. Find the probability that the second battery selected is not defective, given that the first was not defective. Let N i denote the event that the i th battery selected is nondefective. P(N 2 N 1 ) = P(N 1N 2 ) P(N 1 ) where and P(N 1 N 2 ) = 3 2 4 3 = 1 2 P(N 1 ) = 3 4 Therefore P(N 2 N 1 ) = 2 3. Arthur Berg 9/ 11
Example 3.2 Arthur Berg 10/ 11
Three axioms of probability Conditional probabilities satisfy the three axioms of probability. (i) 0 P(A B) 1 (ii) P(S B) = 1 (iii) If A 1, A 2,... are mutually exclusive events, then so are A 1 B, A 2 B,..., and ( ) P A i B = P(A i B) i=1 i=1 Arthur Berg 11/ 11