1. Day 59 – Joint Frequencies

2. Problem
The table below shows the number of M&Ms and Skittles in another class’ candy mixture. Use the data in the table to complete the problems that follow.
RED
ORANGE
YELLOW
GREEN
BLUE
PURPLE
BROWN
TOTAL
M & M’s 12 6 8 14 10 58
Skittle’s 9 11 7 5 49 21 17 15 24
107

3. 1. Marginal Probability a) Write an example event in words.
RED
ORANGE
YELLOW
GREEN
BLUE
PURPLE
BROWN
TOTAL
M & M’s 12 6 8 14 10 58
Skittle’s 9 11 7 5 49 21 17 15 24
107
a) Write an example event in words.
b) Find the value of the probability.

4. 2. Joint Probability a) Write an example event in words.
RED
ORANGE
YELLOW
GREEN
BLUE
PURPLE
BROWN
TOTAL
M & M’s 12 6 8 14 10 58
Skittle’s 9 11 7 5 49 21 17 15 24
107
a) Write an example event in words.
b) Find the value of the probability.

5. 3. Conditional Probability
RED
ORANGE
YELLOW
GREEN
BLUE
PURPLE
BROWN
TOTAL
M & M’s 12 6 8 14 10 58
Skittle’s 9 11 7 5 49 21 17 15 24
107
a) Write an example event in words.
b) Find the value of the probability.

6. Problem
Jessica surveys students at her school about their favorite food. She recorded the responses in the table below.
PIZZA
CHICKEN
STEAK
FRENCH FRIES
TOTALS
MALES 14 10 6 7
FEMALES 12 13 4 8

7. 1. What is the joint frequency of female students who prefer Pizza?
CHICKEN
STEAK
FRENCH FRIES
TOTALS
MALES 14 10 6 7
FEMALES 12 13 4 8

8. 2. What is the joint frequency of male students who prefer Steak?
PIZZA
CHICKEN
STEAK
FRENCH FRIES
TOTALS
MALES 14 10 6 7
FEMALES 12 13 4 8

9. 3. What is the marginal frequency for each type of food?
PIZZA
CHICKEN
STEAK
FRENCH FRIES
TOTALS
MALES 14 10 6 7
FEMALES 12 13 4 8

10. 4. What is the marginal frequency for each gender?
PIZZA
CHICKEN
STEAK
FRENCH FRIES
TOTALS
MALES 14 10 6 7
FEMALES 12 13 4 8

11. 5. Given that the student is female, what is the frequency that she likes french fries?
PIZZA
CHICKEN
STEAK
FRENCH FRIES
TOTALS
MALES 14 10 6 7
FEMALES 12 13 4 8

12. Answer Key
1. a. An example of a student response might be to randomly select an M&M from the classroom “mix” of candy. This is marginal probability because only one characteristic needs to occur. b. Then, the probability of this would be 58/ a. An example of a student response might be to select the event of a purple Skittle selected at random. This is a joint probability because the outcome requires that two characteristics occur at the same time. b. Then, the probability would be 5/ a. An example of a student response might be to pose the question, “If a green candy is selected, what is the probability that it is a Skittle?”. This is a conditional probability because one condition is given, restricting the sample space that needs to be considered. b. Then, the probability would be 10/24, as only the “Green” candy column is considered.

13. 1. 12/37
2. 6/37
3. Pizza. 23/74 = 13/37
Chicken. 23/74 = 13/37
Steak. 10/74 = 5/37
French Fries 15/74
4.Male 37/74 = ½
Female 37/74 = ½
5. 8/37