1. WARM – UP
Auto accidents involve many factors. Examining only two, 25% of accidents involve women, 65% involve rainy conditions, and 16% of accidents involve both rain AND females.
1. Given that it was raining, what is the probability that an accident involves a woman?
2. Given that a woman got into an accident, what is the probability that it was raining?
3. Are the two events Mutually Exclusive, Independent, or Neither? EXPLAIN.
P(W | R) = P(W ∩ R) / P(R)
P(W | R) = 0.16 / 0.65
P(W | R) =
P(R | W) = P(R ∩ W) / P(W)
P(R | W) = 0.16 / 0.25
P(R | W) = 0.64
NEITHER
NOT M.E. because P(W ∩ R) ≠ 0
NOT Independent because P(W ∩ R) = 0.16 ≠ P(W) x P(R) =
OR P(R) = 0.65 ≠ P(R|W) = 0.64

2. 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

3. 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

4. Ch. 15 - Conditional Probability Trees
EXAMPLE: Each morning coffee is prepared for the entire office staff by one of three employees, depending on who arrives first. Debbie arrives first 20% of the time; Karen arrives first 30% of the time; and Amy the remaining percent. The three are not good coffee makers. Debbie has a 0.1 probability of making bad coffee, Karen 0.4, and Amy If YOU arrive to work and find bad coffee, what if the probability that it was prepared by Debbie?
B = 0.1
P(D|B) = ?
P(D|B) = P(D ∩ B) P(B)
D = 0.2 OR
B = 0.4
K = 0.3 OR
P(D ∩ B) = (0.2)(0.1) = 0.02
P(B) = (0.2)(0.1)+(0.3)(0.4)+(0.5)(0.3)
= 0.29
B = 0.3
A = 0.5
P(D|B) = = 0.069

5. P(Bc) = (0.2)(0.9)+(0.3)(0.6)+(0.5)(0.7) = 0.71 or P(Bc) = 1 – P(B)
D = 0.2
Bc = 0.9
B = 0.4 OR
K = 0.3
Bc = 0.6
B = 0.3 OR
A = 0.5
Bc = 0.7
What is the probability that YOU arrive to work and the Coffee is NOT Bitter?
P(Bc) = ?
P(Bc) = (0.2)(0.9)+(0.3)(0.6)+(0.5)(0.7) = 0.71 or
P(Bc) = 1 – P(B)

6. P(Chuck | Break) =
P(Chuck ∩ Break)
P(Break)

7. 0.5625 PAGE 366 #45 P(Chuck | Break) = P(Chuck ∩ Break) P(Break)
(0.3)(0.03)
(0.4)(0.01)+(0.3)(0.01)+(0.3)(0.03)
0.5625

9. PAGE 366
#43

10. = 0.353 = 0.033 = 0.2 a.) P(Detain | Not Drinking) b.) P(Detain) =
P(Drinking ∩ Det.) + (Not Drinking ∩ Det.)
= (0.12)(0.8)+(0.88)(0.2)
= 0.272
c) P(Drunk | Det.)=
P (Drunk ∩ Det.)
P (Detain)
= (0.12)(0.8)
(0.12)(0.8) + (0.88)(0.2)
= 0.353
d) P(Drunk | Release)=
P (Drunk ∩ Release)
P (Release)
= (0.12)(0.2)
(0.12)(0.2) + (0.88)(0.8)
= 0.033

11. HW Page 366: 35, 36,39, 40
If P(Lug|On Time) = P(Lug|Not on Time) INDEPENDENT.

12. HW Page 366: 35, 36, 39, 40

13. If P(B) = P(B|A) then A and B are INDEPENDENT.
(a) No, P(Lug.|On Time)=0.95 ≠ P(Lug.|Not on Time)=0.65

14. PAGE 366
#35 a)
P(On Time)=0.15
P(Lug.|On Time)=0.95
P(Lug.|Not onTime)=0.65
b) P(Luggage)=
P(On time ∩ Luggage)+P(Not on time ∩ Luggage)
=(0.15)(0.95)+(0.85)(0.65)
=0.695

16. 0.467 PAGE 366 #46 P(Supplier A | Defective) =
P(Defective)
(0.7)(0.01)
(0.7)(0.01)+(0.2)(0.02)+(0.1)(0.04)
0.467

17. P(Absent|Day) ≠ P(Absent|Night)
(a) The are NOT independent because…
P(Absent|Day) ≠ P(Absent|Night)

18. P(Day)=0.60
P(Abs.|Day)=0.95
P(Abs.|Night)=0.65
P(Absent)=?

19. P(Cond.|Smoker) ≠ P(Cond.|Nonsmoker)
(a) The are NOT independent because…
P(Cond.|Smoker) ≠ P(Cond.|Nonsmoker)

20. P(Smoke)=0.23
P(Cond.|Smoke)=0.57 P(Cond.|SmokeC)=0.13
P(Lung Cond.)=?

21. WORKSHEET

22. Ch. 15 - Conditional Probability Trees
Conditional Probability = Intersection divided by the Condition.
EXAMPLE: Democrat Republican NO Party
West
Northeast
Southeast 68 57 77
202
Find the Probability of selecting a democrat given you only select from the Northeast.
Find the Probability of selecting a person from the west given you only select from Republicans.