WARM UP Auto accidents involve many factors Examining

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? P(W | R) = P(W ∩ R) / P(R) P(W | R) = 0. 16 / 0. 65 P(W | R) = 0. 2462 2. Given that a woman got into an accident, what is the probability that it was raining? P(R | W) = P(R ∩ W) / P(W) P(R | W) = 0. 64 P(R | W) = 0. 16 / 0. 25 3. Are the two events Mutually Exclusive, Independent, or Neither? EXPLAIN. NEITHER NOT M. E. because P(W ∩ R) ≠ 0 NOT Independent because P(W ∩ R) = 0. 16 ≠ P(W) x P(R) = 0. 1625 OR P(R) = 0. 65 ≠ P(R|W) = 0. 64

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. a. What is the probability that a person selected at random would receive an AP Stat grade of 5? a. ) P(Gets a 5) = ? P(5) = Studied and got a 5 OR Did NOT study and got a 5. P(5) = (0. 60)(0. 88) + (0. 40)(0. 009) P(5) = 0. 5316 Studies 5 = 0. 88 0. 60 5 c = 0. 12 OR 5 =. 009 0. 40 Does NOT Study 5 c =0. 991

P(5) = 0. 5316 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 5 c = 0. 12 OR 5 =. 009 0. 40 Does NOT Study 5 c =0. 991

Ch. 15 - Conditional Probability Trees 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 0. 3. If YOU arrive to work and find bad coffee, what if the probability that it was prepared by B = 0. 1 Debbie? D = 0. 2 P(D|B) = ? OR EXAMPLE: P(D|B) = 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 P(D|B) = = 0. 069 B = 0. 4 K = 0. 3 OR B = 0. 3 A = 0. 5

B = 0. 1 D = 0. 2 Bc = 0. 9 B = 0. 4 OR K = 0. 3 Bc = 0. 6 B = 0. 3 A = 0. 5 OR Bc = 0. 7 What is the probability that YOU arrive to work and the Coffee is NOT Bitter? P(Bc) = (0. 2)(0. 9)+(0. 3)(0. 6)+(0. 5)(0. 7) = 0. 71 or P(Bc) = 1 – P(B)

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

PAGE 366 #45 P(Chuck | Break) = P(Chuck ∩ Break) P(Break) (0. 3)(0. 03) (0. 4)(0. 01)+(0. 3)(0. 03). 0. 5625

PAGE 366 #43

a. ) P(Detain | Not Drinking) = 0. 2 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. 88)(0. 2). = 0. 353 d) P(Drunk | Release)= P (Drunk ∩ Release) P (Release) = (0. 12)(0. 2) + (0. 88)(0. 8). = 0. 033

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

HW Page 366: 35, 36, 39, 40

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

PAGE 366 #35 a) P(On Time)=0. 15 P(Lug. |On Time)=0. 95 P(Lug. |Not on. Time)=0. 65 b) P(Luggage)= P(On time ∩ Luggage)+P(Not on time ∩ Luggage) =(0. 15)(0. 95)+(0. 85)(0. 65) =0. 695

PAGE 366 #46 P(Supplier A | Defective) = P(Supplier A ∩ Defective) P(Defective) (0. 7)(0. 01)+(0. 2)(0. 02)+(0. 1)(0. 04). 0. 467

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

P(Day)=0. 60 P(Abs. |Day)=0. 95 P(Abs. |Night)=0. 65 P(Absent)=?

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

P(Smoke)=0. 23 P(Cond. |Smoke)=0. 57 P(Cond. |Smoke. C)=0. 13 P(Lung Cond. )=?

WORKSHEET

Ch. 15 - Conditional Probability Trees Conditional Probability the Condition. EXAMPLE: West Northeast Southeast Democrat 39 15 30 84 = Intersection divided by Republican NO Party 17 30 31 78 12 12 16 68 57 77 40 202 1. Find the Probability of selecting a democrat given you only select from the Northeast. 2. Find the Probability of selecting a person from the west given you only select from Republicans.