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) = 0.2462 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) = 0.1625 OR P(R) = 0.65 ≠ P(R|W) = 0.64

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) = 0.5316 5 = .009 0.40 Does NOT Study 5c =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 5c = 0.12 OR 5 = .009 0.40 Does NOT Study 5c =0.991

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 0.3. 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

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)

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

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

PAGE 366 #43

= 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

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 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

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

P(Absent|Day) ≠ P(Absent|Night) (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)=?

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

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

WORKSHEET

Ch. 15 - Conditional Probability Trees Conditional Probability = Intersection divided by the Condition. EXAMPLE: Democrat Republican NO Party West 39 17 12 Northeast 15 30 12 Southeast 30 31 16 68 57 77 84 78 40 202 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.