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.
Bell Ringer
Solution (Using a Tree Diagram)
B
B
B
BB
B
B
G
G
G G
G
G
G
BBG
S = {BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG}
CHAPTER 15 PART 1
Probability Rules
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)
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.
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?
𝑃 ( 𝐴∪𝐵 )=𝑃 ( 𝐴 )+𝑃 (𝐵 )− 𝑃 (𝐴∩𝐵)
𝑃 ( 𝐴∪𝐵 )=.56+.62− .42=0.76
Let A = student living on campus and B = student has a meal plan
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.
0.42 0.200.14
A B
0.24
Venn Diagra
m
Conditional Probability𝑷 (𝑩∨𝑨)= 𝒕𝒉𝒆 𝒑𝒓𝒐𝒃𝒂𝒃𝒊𝒍𝒊𝒕𝒚 𝒐𝒇 𝑩 𝒈𝒊𝒗𝒆𝒏 𝑨
𝑷 (𝑩∨𝑨)=𝑷 (𝑨∩𝑩)𝑷 (𝑨)
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 (𝑜𝑛𝑐𝑎𝑚𝑝𝑢𝑠|𝑚𝑒𝑎𝑙𝑝𝑙𝑎𝑛 )= 𝑃 (𝑚𝑒𝑎𝑙𝑝𝑙𝑎𝑛∩𝑜𝑛𝑐𝑎𝑚𝑝𝑢𝑠)
𝑃 (𝑚𝑒𝑎𝑙𝑝𝑙𝑎𝑛)
𝑃 (𝑜𝑛𝑐𝑎𝑚𝑝𝑢𝑠|𝑚𝑒𝑎𝑙𝑝𝑙𝑎𝑛 )=0.420.62
=0.677
Conditional Probability and Independent Events
then events A and B are independent
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?
2. What is the probability that a randomly selected U.S. household owns both a cat and a dog?
3. What is the probability that a randomly selected U.S. household owns a cat if the household owns a 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?
¿1−0.60=0.40
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 = 0.34 + 0.39 – x x=0.13
0.34 0.39 unknown
0.60
𝑃 (𝑐𝑎𝑡∩𝑑𝑜𝑔 )=0.13
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)
=0.130.39
=0.33
Today’s AssignmentRead Chapter 15 Add to HW #9: page 361 #1-4
Chapter 14,15,16 will be included in HW #9 – Due after Thanksgiving Break