1. WARM – UP
The future of an Advertisement Agency depends on the company landing huge clients. For a certain agency, the probability of landing the business of American Airlines is The probability of landing United Airlines is 0.28.
Find the probability of getting at least one client’s business if they are completely Independent.
Find the probability of NOT getting either client’s business if they are completely Independent.
3. Find the probability of getting at least one company’s business if they are Mutually Exclusive events.
HINT: With only 2 events, “At least one” = “P(A OR B) ”
P(A U B) = P(A) + P(B) – P(A ∩ B)
Ind.: P(A ∩ B) = P(A) x P(B)
P(A U B) = – 0.15x0.28
P(A U B) = 0.388
P(Ac ∩ Bc) = 0.612
P(Ac ∩ Bc) = 1 – P(A U B)
P(A U B) = P(A) + P(B) – P(A ∩ B)
M.E. : P(A ∩ B) = 0
P(A U B) = – 0
P(A U B) = 0.43

2. 17
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6. Chapter 15 - CONDITIONAL PROBABILITY
Conditional Probability is the probability of two DEPENDENT events occurring. The P(A and B) is dependent on whether or not event A effects B by occurring first.
Joint Probability: P(A ∩ B) = P(A)·P(B|A)
Conditional Probability: P(B|A) is read,
“The probability of B given A.”
P(B|A) = P(A ∩ B)
P(A)

7. 8 dogs and 6 cats are male. Find the probability that:
EXAMPLE: In a monthly report, an animal shelter states that it currently has 24 dogs and 18 cats available for adoption.
8 dogs and 6 cats are male. Find the probability that:
The pet is male, given that it is a cat.
The pet is a cat, given that it is a female.
The pet is female, given that it is a dog.
42 total pets
P(M|C) = P(M ∩ C)
P(C)
P(C|F) = P(C ∩ F)
P(F)
P(F|D) = P(F ∩ D)
P(D)

8. Conditional Probability Two Way Tables
(Contingency Tables)
EXAMPLE: A psychologist surveyed 478 children in elementary grade school on what their primary goals were in school.
Goals
Sex
Get good Grades
Be Popular
Play Sports
Boys
117 50 60
Girls
130 91 30
227
251
If a student is selected at random, what is the:
1. The Probability the student wants to play sports?
2. The Probability the student wants to be popular given it is a girl?
3. The Probability the student is a girl given we selected a student
wanting to be popular?
P(Sport) = 90/478 = 0.188
P(Pop|Girl) =
P(Pop ∩ Girl) / P(Girl)
= (91/478) / (251/478) = 0.363
P(Girl|Pop) =
P(Girl ∩ Pop) / P(Pop)
= (91/478) / (141/478) = 0.645

9. If P(B) = P(B|A) then A and B are INDEPENDENT.
EXAMPLE: Democrat Republican NO Party
West
Northeast
Southeast 68 57 77
202
Find the Probability of being a Democrat.
Find the Probability of living in the Northeast?
Given that a resident of the Northeast is randomly chosen, what is the probability that person is a Democrat.
Is living in the Northeast independent from being a Democrat? Why or Why not?
P(D) =
P(N) =
P(D|N) = P(D ∩ N)
P(N) =
0.2822 =
NO, b/c P(D|N) ≠ P(D) ≠
If P(B) = P(B|A) then A and B are INDEPENDENT.

10. A recent investigation about the AP Statistics Exam revealed that 88% of students who study get a grade of 5. But 0.9% of the time people who do not study get 5’s. Large-scale studies have shown that 60% of the population actually study .
What is the probability that a person selected at random would receive an AP Stat grade of 5?
b. What is the probability that a person selected at random who received a 5 (given they received a 5), actually studied?

11. P(5) = Studied and got a 5 OR Did NOT study and got a 5.
A recent investigation about the AP Statistics Exam revealed that 88% of students who study get a grade of 5. But 0.9% of the time people who do not study get 5’s. Large-scale studies have shown that 60% of the population actually study .
What is the probability that a person selected at random would receive an AP Stat grade of 5?
a.) P(Gets a 5) = ?
Studies
5 = 0.88
P(5) = Studied and got a 5 OR
Did NOT study and got a 5.
0.60
5c = 0.12 OR
P(5) = (0.60)(0.88) +
(0.40)(0.009)
P(5) =
5 = .009
0.40
Does NOT
Study
5c =0.991

12. P(5) =
b. What is the probability that a person selected at random who received a 5 (given they received a 5), actually studied?
Studies
5 = 0.88
0.60
5c = 0.12 OR
5 = .009
0.40
Does NOT
Study
5c =0.991

13. Homework: PAGE 363: 9 -12, 19 – 20
SEPARATE SHEET OF COLLECTABLE PAPER!!

16. HINTS:

17. P(>$100 | R) = P(R ∩ >$100) P(R)
EXAMPLE: 20% of cars have faulty pollution control systems. If the car needs to be repaired, the cost of repairs exceeds $100 about 40% of the time. When a driver takes in his car for inspection, what is the probability that the car will need to be repaired and he will pay more than $100 for repairs?
P(R ∩ >$100) = ?
P(R) = 0.20
P(>$100 | R) = 0.40
P(>$100 | R) = P(R ∩ >$100)
P(R)
0.40 = P(R ∩ >$100)
0.20
(0.20)(0.40) = 0.08

18. = P(♦)·P(K|♦= not a K) OR P(♦)·P(K|♦=was a K)
EXAMPLE: Two cards are drawn out of a standard deck of cards without replacement. What is the Probability of selecting a Diamond and then a King?
P(♦ ∩ K) = P(♦)·P(K|♦)
= P(♦)·P(K|♦= not a K) OR P(♦)·P(K|♦=was a K)

19. W.U. A sports-drink company surveyed 600 athletes
to find out if they liked Sports Drink G or Sports Drink P. The diagram shows the results of the survey represents ONLY Drink G and 211 represents ONLY P, not both.
Which expression can be used to determine the number of athletes surveyed who did not
like either Sports Drink G or Sports Drink P?
A ( )
B ( )
C 600 − ( )
D 600 − ( )