2. To prove independence you must show the following:
Example 1: Suppose that 60% of all customers of a large insurance agency have automobile policies with the agency (event A), 40% have homeowners policies (event H), and 25% have both. Are events A and H independent? Explain.
To prove independence you must show the following:
For independence P(A) = P(A|B) so
P(auto) = P(auto|home)
0.6 = P(auto∩home)
P(home)
= 0.25
0.4
0.6 ≠ 0.625
Therefore the events are NOT independent
You must show all this work including all the wording that I wrote for full credit!

3. P (D ∪ M) = P(D) + P(M) – P(D ∩ M) = 0.4 + 0.5 – P(D ∩ M)
Example 2: In the ABC Health Club, the probability that a member picked at random is a doctor is .4, the probability that the member is a male is 0.5, and the probability that the member is a male given that he is a doctor is Find the probability that the member is a doctor or a male.
Let D = doctor M = male
P (D ∪ M) = P(D) + P(M) – P(D ∩ M)
= – P(D ∩ M)
To find P(D ∩ M) we will use the formula:
So P(M | D) = P(D ∩ M)
P(D)
0.6 = P(D ∩ M)
0.4
P(D ∩ M) = 0.24
P (D ∪ M) = P(D) + P(M) – P(D ∩ M)
= – 0.24
= 0.66

4. Let D = desktop and L = laptop P(L | D) = P(L ∩ D) P(D) = 0.3 0.8
Example 3: Suppose 80% of the homes in Katy have a desktop computer, 70% have a laptop computer, and 30% have both a desktop computer and a laptop computer. What is the probability that a randomly selected home will have a laptop computer given that they have a desktop computer?
Let D = desktop and L = laptop
P(L | D) = P(L ∩ D)
P(D)
= 0.3
0.8
= 0.375

5. Example 4: Due to turnover and absenteeism at an assembly plant, 20% of the items are assembled by inexperienced employees. Management has determined that customers return 12% of the items assembled by inexperienced employees, whereas only 3% of the items assembled by experienced employees are returned. (hint: use a tree diagram)
a. What is the probability that an item is returned? _______
   
b. If an item is returned, what is the probability that it was assembled by an inexperienced employee? _______
0.048
0.12
Returned
An item that is returned could have been made by an inexperienced worker or an experienced worker
So we do 0.2(.12) + 0.8(.03) = 0.048
0.2
Inexperienced
0.88
Not Returned
0.03
Returned
0.8
Experienced
0.97
Not Returned
0.5
In other words, find the probability an item was assembled by an inexperience worker given it was returned.
P(Inexperienced | Returned) = (Inexperienced ∩ Retuned)
P(Returned)
= 0.2(.12)
0.048
= 0.5

6. P(Karen | Bitter) = P(K ∩ B) P(B) = 0.2(.1)______
Example 5: Each morning coffee is prepared for the entire office staff by one of three employees, depending on who arrives first at work. Karen arrives first 20% of the time; Gail and Larry are each the first to arrive on half of the remaining mornings. The probability that the coffee is bitter when prepared by Karen is 0.1, while the corresponding probabilities when it is prepared by Gail and Larry are .2 and .3, respectively. If you arrive one morning and find the coffee is bitter, what is the probability that it was prepared by Karen? By Larry? What is the probability that it is not bitter? (Hint: use a tree diagram)
In other words, find the probability that Karen made the coffee given that it was bitter:
P(Karen | Bitter) = P(K ∩ B)
P(B)
= (.1)______
0.2(.1) + 0.4(.2) + 0.4(.3)
= 0.09

7. Drink beer Don’t drink beer Smoke 315 165 Don’t Smoke 585 135
Example 6: The results of a survey of drinking and smoking habits of 1200 college students:
Drink beer Don’t drink beer
Smoke
Don’t Smoke
Find the probability of the following:
a. Someone in this group smokes? ________
b. A student smokes given that he drinks? ________
A student drinks given that he smokes? ________
d. Are the two events independent? ________Explain:
Total
480
720
Total
Total total 1200
480/1200
315/900
Only look at the drinking column since that is the given condition. Within the drinking column there are 315/900 who smoke
315/480
Only look at the smoking row since that is the given condition. Within the smoking row there are 315/480 who drink
For independence P(A) = P(A|B) so
P(smoking) = P(smoking|drinks)
480/ = 315/900
0.4 ≠ 0.35
Therefore the events are NOT independent

8. Example 7: Check of dorm rooms on a large college campus revealed that 38% had refrigerators, 52% had TV’s, and 21% had both a TV and a fridge. What’s the probability that a randomly selected dorm room has: (hint: draw a venn-diagram)
a TV but not a fridge? _________
b) a TV or a fridge but not both? _________
c) neither a TV nor a fridge? _________
0.52 – 0.21 = 0.31
= 0.48
1 – 0.31 – 0.21 – 0.17 = 0.31 TV
Fridge
0.31
0.21
0.17
Since it needs to add to
100% then there are some rooms
That don’t have TV’s nor fridges
0.31

9. Example 8: The local animal shelter stated that it currently has 24 dogs and 18 cats available for adoption. Eight of the dogs and 6 of the cats are male. Find each of the following conditional probabilities: (hint: use a table)
the pet is male, given that it is a cat. _________
b) The pet is a cat, given that it is female. _________
c) The pet is female, given that is a dog. _________
6/18
12/28
16/24