See below for an example on gym use in a large apartment complex: Men Women Total Men Women Total Used gym Used gym

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1 Business Stats / Math 1111 Learning Centre Contingency Tables When dealing with probability, it is helpful to think about the sample space when approaching a problem. The sample space is a complete list or representation of all the possible outcomes of an activity. There are many ways that a sample space can be expressed. Sample spaces can be represented with contingency tables, lists, pictures, or tree diagrams. This worksheet focuses on constructing, understanding and using contingency tables. Contingency tables are a way of classifying sample observations according to two or more identifiable characteristics (for example, customer satisfaction and location). Specifically a contingency table shows a count or number of observations grouped by characteristics. A joint probability table shows the same information but converts the counts into a relative frequency or probability. This is done by dividing the observations for a particular characteristic(s) by the total sample size. See below for an example on gym use in a large apartment complex: Contingency Table Joint Probability Table Men Women Men Women Used gym Used gym Did not Did not use use gym gym We read the contingency table as saying that out of 350 residents in an apartment complex, 170 were men, and 180 were women. Of the 170 men, 65 used the gym facilities. Of the 180 women, 145 used the gym facilities. Note that each column is added downwards for a total and each row is added across for a total. Also, the sum of the row totals and the sum of the column totals must equal the entire sample size, which is shown in the lower right corner of the table (350 people). Knowing these two pieces of information allows us to fill in a contingency table when only some of the data is given. In a joint probability table, we take each count or number of observations and divide by the total sample size. For example, men who used the gym: the frequency of this event is 65/350 = 0.19 or a 19% probability. We do this for each value. Now the total frequency (lower right corner of the table) should be Recall that the sum of probabilities of all possible outcomes for an event must equal one. Again the totals for the rows and the totals for the columns when summed much equal the total in the lower right hand corner. Example 1: If the probability of event A is 0.4 {P(A) = 0.40}, the probability of event B {P(B) = 0.10}, and the probability of A and B is 0.03, construct a joint probability table. Authored by Emily Simpson Student review only. May not be reproduced for classes.

2 Solution: The first step is to draw the table, label columns and rows, and enter the data given. We know the lower right corner must be 1.0 and we can fill in the given information (shown in bold): the probability of event A is the total probability for A, no matter what s happening with B. We put that number in the row under column A. Similarly, the probability of event B means the total probability for event B regardless of what happens with A, so it goes in the column on row B. The probability of A and B occurring belongs in the cell of the table where column A and row B intersect. Now we can determine the missing probabilities: P(A) + P(not A) = 1.0 P(not A) = 1.0 P(A) = = 0.60 B P(B) + P(not B) = 1.0 P(not B) = 1.0 P(B) = = 0.90 not B P(B and A) + P(B and not A) = 0.10, so P(B and not A) = = 0.07 P(A and B) + P(A and not B) = 0.40, so P(A and not B) = = 0.37 We know P(not B) = 0.90 and P(not B and A)=0.37, so P(not B and not A) = = 0.53 We could also solve for P(not B and not A) using P(not A and B) + P(not A and not B) = P(not A), so P(not B and not A) = P(not A) P(not A and B) = = 0.53 We can also derive probabilities for events using information from the table and the probability rules. Example 2: If one person is picked randomly from the 350 apartment residents, what is the probability of picking a woman? Solution: In this case, we don t care if the woman used the gym facilities or not, we simply want the total probability for choosing a woman so we go to the total at the bottom of the Woman column: P(Woman) = 0.51 Example 3: What is the probability of picking a woman and that the woman uses the gym facilities? Solution: This can be expressed as P(Woman and Used Gym). Here we go to the cell in the table in the column for Woman and the row for Used gym = 0.41 Example 4: What is the probability that a person chosen at random used the gym GIVEN that the person was a woman? Solution: This is a slightly different case called conditional probability where we want to know the probability of a event, GIVEN that something else is true/has happened. The formula for this is: We read this as the probability of event A, given event B is equal to the probability of A AND B occurring divided by the probability of B occurring. Student review only. May not be reproduced for classes. 2

3 We would rewrite our question in mathematical form as: For more review, see the Stats worksheet on Conditional Probability. Two other pieces of information that may affect your use or construction of a joint probability table are (1) if two events are mutually exclusive, then the probability of both events occurring at the same time is zero {P(A and B) = 0}, and (2) if two events are independent, if the probability of one event occurring is unaffected by whether the other event has occurred. We would express this mathematically as: EXERCISES Round all answers to two decimal places. 1. If P(A) = 0.24, P(B) = 0.30, and if P(A and not B) = 0.14, find: P(not B), P(A and B), P(B or not A), P(B A), and P(not A not B). 2. If P(A) = 0.40 and P(B) = 0.23, and A and B are mutually exclusive events, find: P(not A), P(not B and not A), P(not A or B), P(A B), and P(not B not A). 3. If P(A) = 0.3 and P(not B) = 0.40, and if A and B are independent, find: P(B), P(A or B), P(not A and B), P(not A not B), and P(B not A). 4. If P(A) = 0.16 and P(B) = 0.48, and P(A B) = 0.3, find: P(not A), P(A or not B), P(A and B), P(not A not B), and P(B A). 5. In a random sample of 250 students commuting to Vancouver Community College, 190 use Translink (transit), and 130 of those are coming from a distance greater than 15 km away. Of those who do not use transit, 40 are commuting from less than 15 km away. (a) What is the probability that a person in the sample would commute more than 15 km to VCC? (b)what is the probability that a person would be taking transit and commuting less than 15 km? (c) What is the probability that a person would not be taking transit or would be commuting more than 15 km? (d) What is the probability of a person not taking transit given that he/she is commuting less than 15 km? (e) What is the probability of a person commuting more than 15 km given that she/he is taking transit? 6. You decide to take a survey of customers at your new café to determine whether there is any difference in the types of drink ordered depending on the sex of the customer. The three categories of interest are hot coffee drinks, cold coffee drinks, and all other beverages. Student review only. May not be reproduced for classes. 3

4 A total of 350 customers are surveyed; of those, 160 are women. 115 customers order a cold coffee beverage and 76 of those customers are men. Ninety-nine customers ordered other beverages and 56 women ordered a hot coffee drink. *Use 4 decimal places for these problems* (a) What is the probability that a person in the survey would not be a woman? (b) What is the probability that a customer would order a hot coffee beverage and be a man? (c) What is the probability that a customer would not order a cold coffee drink or be a woman? (d) What is the probability that a cold coffee drink will be ordered given that the customer is a man? (e) What is the probability that the customer is not a man given that an other beverage is ordered? (f) Is the selection of a coffee-based drink independent of customer sex? SOLUTIONS 1. P(not B) = 0.70 P(A and B) = 0.10 P(B or not A) = = 0.86 P(B A) = 0.10/0.24 = 0.42 P(not A not B) = 0.56/0.70 = 0.80 B not B P(not A) = 0.60 P(not B and not A) = 0.37 P(not A or B) = = 0.60 P(A B) = 0 P(not B not A) = 0.37/0.60 = 0.62 B not B P(B) = 0.60 P(A or B) = = 0.72 P(not A and B) = 0.42 P(not A not B) = 0.28/0.40 = 0.70 P(B not A) = 0.42/0.70 = 0.60 B not B P(not A) = 0.84 P(A or not B) = = 0.66 P(A and B) = 0.14 P(not A not B) = 0.50/0.52 = 0.96 P(B A) = 0.14/0.16 = 0.88 B not B Student review only. May not be reproduced for classes. 4

5 5. P(>15km) = 0.60 P(transit and <15 km) = 0.24 P(not transit or >15 km) = = 0.76 P(not transit <15 km) = 0.16/0.40 = 0.40 P(>15km transit) = 0.52/0.76 = 0.68 BC transit Not transit <15km (0.24) (0.16) (0.40) >15km (0.52) (0.08) (0.60) (0.76) (0.24) (1.0) *probabilities are shown in parenthesis 6. Hot coffee Cold coffee drink drink Other Men 80 (0.2286) 76 (0.2171) 34 (0.0971) 190 (0.5429) Women 55 (0.1571) 39 (0.1114) 66 (0.1886) 160 (0.4571) 135 (0.3857) 115 (0.3286) 100 (0.2857) 350 (1.0) (a) P(customer is not a woman) = P(customer is a man) = (b) P(hot coffee drink and man) = (c) P(not cold coffee drink or woman) = (d) P(cold coffee drink man) = (e) P(not man other bev) = (f) You can solve this problem many ways. For example, by evaluating? P(coffee drink man) = P(coffee drink) P(coffee drink man) = ( )/ = P(coffee drink) = = Therefore the selection of coffee-based drinks is NOT independent of customer sex. Student review only. May not be reproduced for classes. 5