1. Bell Ringer
Find the sample space for the gender of the children if a family has three children. Use B for boy and G for girl. Include order. For example, BBG and BGB are different outcomes.

2. Solution (Using a Tree Diagram)B B G
BBG B B G G B
S = {BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG} B G G B G G

3. Chapter 15 part 1
Probability Rules

4. Addition Rule 𝑷 𝑨∪𝑩 =𝑷 𝑨 +𝑷 𝑩 −𝑷(𝑨∩𝑩) Note:
If A and B are disjoint, we just use P(A) + P(B)

5. A survey of college students found that 56% live in a campus residence hall, 62% participate in a campus meal program, and 42% do both.
Let A = student living on campus and B = student has a meal plan
Are living on campus and having a meal plan independent? Are they disjoint?
They are independent, but they are not disjoint.

6. A survey of college students found that 56% live in a campus residence hall, 62% participate in a campus meal program, and 42% do both. What’s the probability that a randomly selected student either lives or eats on campus?
Let A = student living on campus and B = student has a meal plan
𝑃 𝐴∪𝐵 =𝑃 𝐴 +𝑃 𝐵 −𝑃(𝐴∩𝐵)
𝑃 𝐴∪𝐵 = −.42=0.76

7. A survey of college students found that 56% live in a campus residence hall, 62% participate in a campus meal program, and 42% do both. A B
Venn
Diagram
0.14
0.42
0.20
0.24

8. Conditional Probability
𝑷 𝑩|𝑨 ="𝒕𝒉𝒆 𝒑𝒓𝒐𝒃𝒂𝒃𝒊𝒍𝒊𝒕𝒚 𝒐𝒇 𝑩 𝒈𝒊𝒗𝒆𝒏 𝑨"
𝑷 𝑩|𝑨 = 𝑷(𝑨∩𝑩) 𝑷(𝑨)

9. From before, 56% of students live on campus, 62% have meal plans, 42% do both. What is the probability that someone with a meal plan is also living on campus?
𝑃 𝑜𝑛 𝑐𝑎𝑚𝑝𝑢𝑠 𝑚𝑒𝑎𝑙 𝑝𝑙𝑎𝑛 = 𝑃(𝑚𝑒𝑎𝑙 𝑝𝑙𝑎𝑛∩𝑜𝑛 𝑐𝑎𝑚𝑝𝑢𝑠) 𝑃(𝑚𝑒𝑎𝑙 𝑝𝑙𝑎𝑛)
𝑃 𝑜𝑛 𝑐𝑎𝑚𝑝𝑢𝑠 𝑚𝑒𝑎𝑙 𝑝𝑙𝑎𝑛 = =0.677

10. Conditional Probability and Independent Events
𝑰𝒇 𝑷 𝑩|𝑨 =𝑷 𝑩 ,
then events A and B are independent

11. According to a pet owners survey, 39% of U. S
According to a pet owners survey, 39% of U.S. households own at least one dog and 34% of U.S. households own at least one cat. Assume that 60% of U.S. households own a cat or a dog.
What is the probability that a randomly selected U.S. household owns neither a cat nor a dog?
What is the probability that a randomly selected U.S. household owns both a cat and a dog?
What is the probability that a randomly selected U.S. household owns a cat if the household owns a dog?

12. 𝑃 𝑛𝑒𝑖𝑡ℎ𝑒𝑟 𝑐𝑎𝑡 𝑛𝑜𝑟 𝑑𝑜𝑔 =1 −𝑃 𝑐𝑎𝑡∪𝑑𝑜𝑔
According to a pet owners survey, 39% of U.S. households own at least one dog and 34% of U.S. households own at least one cat. Assume that 60% of U.S. households own a cat or a dog.
1. What is the probability that a randomly selected U.S. household owns neither a cat nor a dog?
𝑃 𝑛𝑒𝑖𝑡ℎ𝑒𝑟 𝑐𝑎𝑡 𝑛𝑜𝑟 𝑑𝑜𝑔 =1 −𝑃 𝑐𝑎𝑡∪𝑑𝑜𝑔
=1 −0.60=0.40

13. According to a pet owners survey, 39% of U. S
According to a pet owners survey, 39% of U.S. households own at least one dog and 34% of U.S. households own at least one cat. Assume that 60% of U.S. households own a cat or a dog.
2. What is the probability that a randomly selected U.S. household owns both a cat and a dog?
P cat∪𝑑𝑜𝑔 =𝑃 𝑐𝑎𝑡 +𝑃 𝑑𝑜𝑔 −𝑃(𝑐𝑎𝑡∩𝑑𝑜𝑔)
0.60
unknown
0.34
0.39
= – x → x=0.13
𝑃 𝑐𝑎𝑡∩𝑑𝑜𝑔 =0.13

14. 𝑃 𝑐𝑎𝑡 𝑑𝑜𝑔 = 𝑃(𝑐𝑎𝑡∩𝑑𝑜𝑔) 𝑃(𝑑𝑜𝑔) = 0.13 0.39 =0.33
According to a pet owners survey, 39% of U.S. households own at least one dog and 34% of U.S. households own at least one cat. Assume that 60% of U.S. households own a cat or a dog.
3. What is the probability that a randomly selected U.S. household owns a cat if the household owns a dog?
𝑃 𝑐𝑎𝑡 𝑑𝑜𝑔 = 𝑃(𝑐𝑎𝑡∩𝑑𝑜𝑔) 𝑃(𝑑𝑜𝑔) = =0.33

15. Today’s Assignment
Read Chapter 15
Add to HW #9: page 361 #1-4 Chapter 14,15,16 will be included in HW #9 – Due after Thanksgiving Break