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.
Solution (Using a Tree Diagram)
B
B
G
G
B
G
B
G
B
G
B
G
B
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?
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 = .56 + .62 − .42 = 0.76
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
0.14
B
0.42
Venn
Diagram
0.20
0.24
Conditional Probability
𝑷 𝑩|𝑨 = "𝒕𝒉𝒆 𝒑𝒓𝒐𝒃𝒂𝒃𝒊𝒍𝒊𝒕𝒚 𝒐𝒇 𝑩 𝒈𝒊𝒗𝒆𝒏 𝑨"
𝑷(𝑨 ∩ 𝑩)
𝑷 𝑩|𝑨 =
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(𝑚𝑒𝑎𝑙 𝑝𝑙𝑎𝑛 ∩ 𝑜𝑛 𝑐𝑎𝑚𝑝𝑢𝑠)
𝑃 𝑜𝑛 𝑐𝑎𝑚𝑝𝑢𝑠 𝑚𝑒𝑎𝑙 𝑝𝑙𝑎𝑛 =
𝑃(𝑚𝑒𝑎𝑙 𝑝𝑙𝑎𝑛)
0.42
𝑃 𝑜𝑛 𝑐𝑎𝑚𝑝𝑢𝑠 𝑚𝑒𝑎𝑙 𝑝𝑙𝑎𝑛 =
= 0.677
0.62
Conditional Probability and
Independent Events
𝑰𝒇 𝑷 𝑩|𝑨 = 𝑷 𝑩 ,
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?
𝑃 𝑛𝑒𝑖𝑡ℎ𝑒𝑟 𝑐𝑎𝑡 𝑛𝑜𝑟 𝑑𝑜𝑔 = 1 − 𝑃 𝑐𝑎𝑡 ∪ 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
0.60
0.34
=
0.39
0.34 + 0.39 – x
unknown
→
𝑃 𝑐𝑎𝑡 ∩ 𝑑𝑜𝑔 = 0.13
x=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)
0.13
𝑃 𝑐𝑎𝑡 𝑑𝑜𝑔 =
=
= 0.33
𝑃(𝑑𝑜𝑔)
0.39
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