Probability The Study of Chance! Part 2

In this powerpoint we will continue calculating probabilities using In this powerpoint we will continue calculating probabilities using Venn DiagramsVenn Diagrams Contingency TablesContingency Tables

Let’s begin with Venn Diagrams A venn diagram consists of interlocking circles that allows us to see relationships Learn about the history of the Venn diagram at #venn #venn A great website created and maintained by a great educator, Pat Ballew.

Relationships with Two Outcomes Consider for a moment the following: Consider for a moment the following: At the Pueblo Athletic Club, 72% of the members use the nautilus machines, 51% use the pool, and 30% use both. Create a venn diagram that shows this relationship. Step 1: Since there are 2 possibilities, we will use a venn diagram with 2 circles. Step 2: Now begin by putting the proportion of people who use both the Nautilus machines and the swimming pool. P(M and S) =.30 Step 3: Notice that the circle for the nautilus machines already has the 30% of the people who use both. This means that we need to take the total of 72% that use the nautilus machines and subtract those who use both (30%) to find the percent that use the machines but not the pool. P(M and S c ) = =.42 Step 4: We need to do the same process to find the percent of people who use the pool but not the machines. P(M c and S) = =.21 Step 5: To complete the diagram we need to find the percent of people who use neither the machines or the pool. P(M c and S c ) = 1 – ( ) =.07 Nautilus Machines Swimming Pool

Using the Venn diagram to answer questions. Once we have created the diagram, we can answer many different questions. Once we have created the diagram, we can answer many different questions. 1.Find the probability that a randomly selected member uses only the swimming pool Find the probability that a member uses either the machines or the swimming pool, but not both =.63 Nautilus Machines Swimming Pool Find the probability that a member uses either the machines or the swimming pool. In statistics this type of “or” is an inclusive or, which means one or the other or both: so… =.93

Using a Contingency Table We can represent the same situation in a contingency (two-way) table. We can represent the same situation in a contingency (two-way) table. Step 1: Fill in the information we have been given. Step 1: Fill in the information we have been given. Step 2: Since we are dealing with proportions we know the “grand total” is equal to 1. Step 2: Since we are dealing with proportions we know the “grand total” is equal to 1. Step 3: Now using subtraction, we can find the missing proportions. Step 3: Now using subtraction, we can find the missing proportions. Total No Yes TotalNoYesSW I P M O I L NG Nautilus Machines

Using the table to answer questions Since our table is already in proportions we can answer many questions without any calculations What proportion of the membership does not use the Nautilus Machine: P(M c ) =.28 What proportion of the membership uses neither the Nautilus Machines nor the pool? P(M c and S c ) =.07 Total No Yes TotalNoYesSW I P M O I L NG Nautilus Machines

Calculating “or” probabilities Total No Yes TotalNoYesSW I P M O I L NG Nautilus Machines Remember that the use of “or” in statistics is inclusive. The easiest way to find this probability is to draw some lines on the table. First, a line across the row where the member uses the swimming pool. Then, a line down the column where the member uses the Nautilus Machines. Now add the row total and the column total, but notice that the value where the lines intersect has been added in twice. So we must subtract one of them =.93

Using Counts What if our data is in counts rather than proportions? What if our data is in counts rather than proportions? How does this change our calculations? How does this change our calculations? Well let’s see. Well let’s see. In a large shopping mall, a marketing agency conducted a survey on credit cards. Respondents were asked to identify if they were employed or unemployed and whether or not they owned a credit card. The agency found the following. Of the 109 respondents 46 owned a credit card, 62 were unemployed and 18 both owned a credit card and were employed.In a large shopping mall, a marketing agency conducted a survey on credit cards. Respondents were asked to identify if they were employed or unemployed and whether or not they owned a credit card. The agency found the following. Of the 109 respondents 46 owned a credit card, 62 were unemployed and 18 both owned a credit card and were employed.

Counts instead of proportions Total No Yes TotalNoYesJOB Credit Card The agency found the following. Of the 109 respondents 46 owned a credit card, 62 were unemployed and 18 both owned a credit card and were employed.The agency found the following. Of the 109 respondents 46 owned a credit card, 62 were unemployed and 18 both owned a credit card and were employed. Now, using subtraction complete the table.Now, using subtraction complete the table. Find the probability that a randomly selected respondent was employed: P(E) = 47/109Find the probability that a randomly selected respondent was employed: P(E) = 47/109 Find the probability that a randomly selected respondent was both employed and did not have a credit card. P(E and C c )=29/109Find the probability that a randomly selected respondent was both employed and did not have a credit card. P(E and C c )=29/109 Find the probability that a randomly selected respondent either has a job or does not have a credit card. Again for this question, let’s use lines on the table. So, P(E or C c ) = ( – 29) / 109

Three Events??? What if we have three events? What if we have three events? How can we show the relationships? How can we show the relationships? Once again, we will use Venn diagrams to show the relationships. Once again, we will use Venn diagrams to show the relationships.

Three Events In a survey carried out in a school snack shop, the following results were obtained. Of 100 boys questioned, 78 liked sweets, 74 liked ice cream, 53 liked cake, 57 liked both sweets and ice cream, 46 liked both sweets and cake, 10 liked only ice cream, and only 31 liked all three. We’ll create a Venn diagram to model the relationship. In a survey carried out in a school snack shop, the following results were obtained. Of 100 boys questioned, 78 liked sweets, 74 liked ice cream, 53 liked cake, 57 liked both sweets and ice cream, 46 liked both sweets and cake, 10 liked only ice cream, and only 31 liked all three. We’ll create a Venn diagram to model the relationship.

Creating the Venn diagram Start with the intersection of all three circles: 31 liked all three Then move to the intersection two events. 57 liked both sweets and ice cream. So 57-31=26 46 liked both sweets and cake. So = 15 Now look at the single events 10 liked only ice cream 74 liked ice cream total, 74-( ) = 7 53 liked cake, 53 – ( )= liked sweets 78 – ( ) = 26 Notice that this information tells us that all boys like at least one of the three. 15 SweetsIce Cream Cake

Additional Resources The Practice of Statistics—YMM The Practice of Statistics—YMM Pg Pg The Practice of Statistics—YMS The Practice of Statistics—YMS Pg Pg