4/22/2012 11:54:00 PM Lecture 8 Here are some examples that demonstrate Rules 4 and 4a from last lecture  if an exam has 12 multiple choice questions, then that’s 12 questions that are either right or wrong  if there is probability that a student makes fewer than 3 mistakes on the exam is .48 and the probability that a student makes from 3-8 inclusive is .30 Find the probability that a student makes: - More than 8 mistakes - 3 or more mistakes - at most 8 mistakes Which 2 of these three are complementary? Why? First, write it out and read carefully. - We know that: o Less than 3 wrong answers (0, 1, and 2 wrong) has a probability of .48, we also know that 3-8 wrong answers (3, 4, 5, 6, 7, and 8 wrong answers) has a probability of .30. Then write out what we don’t know or what we are trying to figure out. Lets try to find the probability of more than 8 mistakes - More than 8 mistakes means (9, 10, 11, and 12 mitakes) o Here we list the possible outcomes Then to figure out probability…. Add the probability of less than 3 mistakes and 3-8 mistakes and subtract from 1 In this case we end up with a probability of .22 Now lets find the probability of a student getting 3 or more mistakes .30 + .22=.52 ( Note: they are mutually exclusive, a student gets ONE score on each test) Next lets find the probability of a student scoring more than 8 mistakes .48 + .30= .78 Associations in Categorical Variables ANDFull Time Part Time Unemployed Other White 635 138 85 515 Black 137 36 19 80 Hispanic 134 29 16 58 Other 45 8 4 27  Consider the Table. Some theories from last lecture can be summarized in a question like this: what percent of white people work part time? - We would solve it like this (138/685 +138+ 88+ 515) * 100 = 10.0% - 138 is the number of white people that work part time and the denominator is the total number of white people that were in the study heres another example using AND what is the probability of a randomly selected person to be white AND working part time? 138/1969 * 100=7% Conditional probabilities the calculation above is a conditional probability  the conditional probability is a subset of attributes in the particular group being studied or focused on. Using A Given  what is the chance that a Hispanic person is employed full time? We calculate FT| Hisp : 134 / 134+ 29 +16 + 58 * 100 =56.6% This probability of being employed full time is conditioned (conditional probability) of the characteristic of being Hispanic. Table + Formula What is the chance that a person is employed full time?Using a forumula: P (A) (B)= P(A and B) / P(B) Application P(B) = P (Hisp)=134 + 29 +16 + 58 / 1969 = .1204 P(Ft and Hisp) = 134 / 1969 = .0681 P(Ft | Hisp) = P (.0681) / P (.1204) * 100= 56.5% If you don’t have a table you must use the formula method Rule 5a and 5b: Rule 5a: P(A | B) = P (A and B) / P (B) Rule 5b: rearranging the formula using algebra gives two forms which are true because interchanging A and B matter sometimes (shown above). BEWARE - P(B|A) does not equal P(A|B) - P(B|A)= 1/P(A|B) - Association  what is the chance a white person is working full time? P(B)=P(White)=635+138+88+515/1969 =.6988 P(A and B) = 635/ 1969 = .3225 Ft| White: .3225 / .6988 *100 =46.2%  stats professionals would say that a respondents race is associated or related to their employment status Independence  when events aren’t associated (unrelated) they are independent 2 events are independent if knowledge about one event reveled nothing about the other event formula: P (A|B) = P(A) Sequences of Events  by examining the probability of events occurring in a certain order we can determine whether or not the events are independent or dependent, and if the events are associated. If events are associated (Rule 5b) can be used P(A and B) = P(B) P(A|B) However if the events are dependent we have a new rule The multiplication rule (5c)  if A and B are independent events then P(A and B)=P(A) P(B) - If events are independent it is easier to compute them o For example: rolls of a dice o Contrast with Dependence dependence and conditional probabilities go