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.

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