1. Probability #1 Manufacturer produces a large number of remote controlled cars. From past experience, the manufacturer knows that approximately 3% are defective. In a quality control procedure, we randomly select 25 cars for testing. We want to determine various probabilities regarding how many cars are defective. Remember to show your formulas when appropriate. A)Define the random variable of interest, X, in this situation. B) Decide which probability distribution this situation models and show why. Let X be the # of defective RC cars found. B(25, 0.03) Bernoulli Trial because… 1)Only 2 outcomes – defective, not defective 2) Probability of success is constant at 0.03 3) Since the cars were selected at random, each car selected is independent of the others. Were looking for the # of defective RC cars out of 25

2. Probability #1 C) Determine the probability that exactly one of the cars is defective. D) Find the probability that at most two of the cars are defective. B(25, 0.03) P(X = 1) = binompdf(25,.03, 1) 0.361 binomcdf(25,.03, 2) 0.962 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

3. Probability #1 E) In the long run, how many cars would you expect to be defective in a sample such as this? F) What would the standard deviation be for the variable X? Binom(25, 0.03) E(X) = E(X) = = np 0.75 defective cars SD(X) =

4. Probability #2 Winthrop, a pre-med student, has been rejected by all 126 U.S. medical schools. Desperate, he sends his transcripts and MCATs to the two least selective foreign schools he can find --- the branch campuses (East & West) of Swampwater Tech. Based on the success his friends have had there, he estimates that his probability of being accepted at the East campus is 0.7 while his chance of being selected at West Campus is 0.4. He also suspects there is a 75% chance that at least one of his applications will be rejected. What is the probability that at least one of the schools will accept him?

5. Probability #2 Based on the success his friends have had there, he estimates that his probability of being accepted at the East campus is 0.7 while his chance of being selected at West Campus is 0.4. He also suspects there is a 75% chance that at least one of his applications will be rejected. What is the probability that at least one of the schools will accept him? P(At least one rejection) = 0.75 = 1 – P(No Rejections) = 0.75 P(No Rejections) = 0.25

6. Probability #2 Based on the success his friends have had there, he estimates that his probability of being accepted at the East campus is 0.7 while his chance of being selected at West Campus is 0.4. He also suspects there is a 75% chance that at least one of his applications will be rejected. What is the probability that at least one of the schools will accept him? 0 1 2 East Campus AcceptedRejected Accepted Rejected West Campus

7. Probability #2 Based on the success his friends have had there, he estimates that his probability of being accepted at the East campus is 0.7 while his chance of being selected at West Campus is 0.4. He also suspects there is a 75% chance that at least one of his applications will be rejected. What is the probability that at least one of the schools will accept him? 0 1 2 East Campus AcceptedRejected Accepted Rejected West Campus

8. Probability #2 Based on the success his friends have had there, he estimates that his probability of being accepted at the East campus is 0.7 while his chance of being selected at West Campus is 0.4. He also suspects there is a 75% chance that at least one of his applications will be rejected. What is the probability that at least one of the schools will accept him? 0 1 2 East Campus AcceptedRejected Accepted Rejected West Campus

9. Probability #2 Based on the success his friends have had there, he estimates that his probability of being accepted at the East campus is 0.7 while his chance of being selected at West Campus is 0.4. He also suspects there is a 75% chance that at least one of his applications will be rejected. What is the probability that at least one of the schools will accept him? P(No Rejections) =.25.70 East Campus AcceptedRejected Accepted Rejected West Campus.40.25.30.60.45.15 1.00 Click on blue box to reveal cell. 0 1 2

10. Probability #2 Based on the success his friends have had there, he estimates that his probability of being accepted at the East campus is 0.7 while his chance of being selected at West Campus is 0.4. He also suspects there is a 75% chance that at least one of his applications will be rejected. What is the probability that at least one of the schools will accept him? P(No Rejections) =.25.25.45.15 EASTWEST

11. Probability #2 What is the probability that at least one of the schools will accept him?.70 East Campus AcceptedRejected Accepted Rejected West Campus.40.25.30.60.45.15 1.00.25.45.15.25 +.15 +.45 0.85 He has an 85% chance of being accepted to at least one of the school. EASTWEST

12. Probability #3 P(sum is 10 | sum > 8) Suppose that two fair die are tossed. What is the probability that the sum equals ten given that it exceeds 8? P(A | B) = 123456 1234567 2345678 3456789 45678910 56789 11 6789101112 3 10

13. Probability #4 Sweet Sassy Molassey U is playing Backwater A&M for the conference football championship. If Backwater’s 1st string QB is healthy, they have a 70% chance of winning. If they have to start the backup, their chances of winning drop to 30%. The team physician says there is a 80% chance that the first string QB will play.

14. Probability #4 1 st QB Plays Sweet Sassy Molassey U is playing Backwater A&M for the conference football championship. If Backwater’s 1st string QB is healthy, they have a 70% chance of winning. If they have to start the backup, their chances of winning drop to 30%. The team physician says there is a 80% chance that the first string QB will play..80.20 1 st QB Doesn’t Play A&M Win.70.30 A&M Lose A&M Win.30.70 A&M Lose

15. Probability #4 1 st QB Plays.80.20 1 st QB Doesn’t Play A&M Win.70.30 A&M Lose A&M Win.30.70 A&M Lose

16. Probability #4 a. What is the probability that Backwater will win the game? P(1 st QB and they win) = P(2 nd QB and they win) = (.8)(.7) = (.2)(.3) =.56.06.56 +.06 =P(A&M Wins) = 0.62

17. Probability #4 b. Suppose you missed the game but read the next day that Backwater won. What is the probability that the 2nd string QB started? P(A&M Wins) =.62 P(2 nd QB | A&M Won) P(A | B) = =.06.62 0.097.06.56

18. Probability #5 If a person is vaccinated properly, the probability of his/her getting a certain disease is 0.05. Without a vaccination, the probability of getting the disease is 0.35. Assume that 1/3 of the population is properly vaccinated. Vaccinated 1/3 2/3 Not Vaccinated Get Disease.05.95 No Disease Get Disease.35.65 No Disease

19. Probability #5 A) If a person is selected at random from the population, what is the probability of that person’s getting the disease? P(Vaccine & Disease) = P(No Vaccine & Disease) =.017.233 0.25

20. Probability #5 A) If a person is selected at random from the population, what is the probability of that person’s getting the disease? P(Vaccine & Disease) = P(No Vaccine & Disease) =.017.233 0.25

21. Probability #5 B) If a person gets the disease, what is the probability that he/she was vaccinated?.017.233 P(Vaccinated | Get Disease) P(A | B) =.017.25 0.068

22. Probability #6 The probability that a football player weighs more than 230 pounds is 0.69, that he is at least 75 inches tall is 0.55, and that he weighs more than 230 pounds and is at least 75 inches tall is 0.43. Find the probability that he is at least 75 inches tall if he weighs more than 230 pounds.

23. Probability #6 The probability that a football player weighs more than 230 pounds is 0.69, that he is at least 75 inches tall is 0.55, and that he weighs more than 230 pounds and is at least 75 inches tall is 0.43. Find the probability that he is at least 75 inches tall if he weighs more than 230 pounds..45 Height < 75 in.> 75in. < 230 > 230 Weight.31.19.55.69.26.12.43 1.00 Click on blue box to reveal cell.

24. Probability #6 The probability that a football player weighs more than 230 pounds is 0.69, that he is at least 75 inches tall is 0.55, and that he weighs more than 230 pounds and is at least 75 inches tall is 0.43. Find the probability that he is at least 75 inches tall if he weighs more than 230 pounds..43.26.12.29 >230 >75

25. Probability #6 The probability that a football player weighs more than 230 pounds is 0.69, that he is at least 75 inches tall is 0.55, and that he weighs more than 230 pounds and is at least 75 inches tall is 0.43. Find the probability that he is at least 75 inches tall if he weighs more than 230 pounds. P( > 75 | >230).43.69 0.623 P(A | B) = >230>75

26. Probability #7 A bank approves 30% of its personal loan applications. If loan manager Ima Fraid Knot receives 25 loan applications on a given day, what is the probability that: a) she approves 5 or more B(25,.3)Let X be the # of approved loans P(X > 5) = 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 0.090 0.91 binomcdf(25,.3, 4)

27. Probability #7 A bank approves 30% of its personal loan applications. If loan manager Ima Fraid Knot receives 25 loan applications on a given day, what is the probability that: b) she approves fewer than 3 P(X < 3) = 0.009 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 B(25,.3) P(X < 2) binomcdf(25,.3, 2)

28. Probability #7 A bank approves 30% of its personal loan applications. If loan manager Ima Fraid Knot receives 25 loan applications on a given day, what is the probability that: c) she approves the first application G(.3) P(Y = 1) = 0.3 Let Y be position of 1 st approval geometpdf(.3, 1)

29. Probability #7 A bank approves 30% of its personal loan applications. If loan manager Ima Fraid Knot receives 25 loan applications on a given day, what is the probability that: d) Ima goes through more than 5 applications before approving the first one P(Y > 5) = 0.832 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 G(.3) P(Y < 5) = 0.168 geometcdf(.3,5)

30. Probability #8 Police find that a patrol unit gets a 3% arrest record when it sets up a checkpoint for drunk drivers. Find the probability that of 200 drivers checked, there will be exactly a) one arrest. B(200,.03)Let X be the # of arrests P(X = 1) = 0.014 binompdf(200,.03, 1)

31. Probability #8 Police find that a patrol unit gets a 3% arrest record when it sets up a checkpoint for drunk drivers. Find the probability that of 200 drivers checked, there will be exactly b) More than 3 arrests? B(200,.03) P(X > 3) = 0.147 0 1 2 3 4 5 6 7 8 9 10 … 200 0.853 binomcdf(200,.03, 3)

32. Probability #8 Police find that a patrol unit gets a 3% arrest record when it sets up a checkpoint for drunk drivers. c) What is the expected number of arrests? The standard deviation of arrests? Binom(200,.03) E(X) = E(X) = = np (200)(0.03) = 6 SD(X) = The expected number of arrests is 6, with a standard deviation of 2.412 arrests.

33. Probability #8 Police find that a patrol unit gets a 3% arrest record when it sets up a checkpoint for drunk drivers. d) What is the probability that the third driver is the first one arrested? Geom(.03) P(Y = 3) = 0.028 Let Y be position of 1 st arrest geometpdf(.03, 3)

34. Probability #9 The weight of potato chips in a medium size bag is stated to be 10 ounces. The amount that the packaging machine puts in these bags is believed to have a normal distribution with mean 10.1 ounces and standard deviation 0.15 ounces.

35. Probability #9 The weight of potato chips in a medium size bag is stated to be 10 ounces. The amount that the packaging machine puts in these bags is believed to have a normal distribution with mean 10.1 ounces and standard deviation 0.15 ounces. A) What proportion of all bags sold will be underweight? -0.667 0.252 < normalcdf(-99,-.667,0,1)

36. Probability #9 The weight of potato chips in a medium size bag is stated to be 10 ounces. The amount that the packaging machine puts in these bags is believed to have a normal distribution with mean 10.1 ounces and standard deviation 0.15 ounces. B) The bags are packed in cases of 24. What is the probability that the mean weight of a 24-bag case will be below 10 ounces? 0.031 0.0006 < z = -3.226 normalcdf(-99,-3.226,0,1)

37. Probability #10 A manufacturing process is designed to produce bolts with a 0.5 inch diameter. Once a day, a random sample of 36 bolts is selected and the diameter recorded. If the resulting sample mean is less than 0.49 inches or greater than 0.51 inches, the process is shut down for adjustment. The standard deviation for the diameter of a single bolt is 0.03 inches. What is the probability that the sample mean fall outside the 0.49 to 0.51 inch interval?

38. Probability #10 A manufacturing process is designed to produce bolts with a 0.5 inch diameter. Once a day, a random sample of 36 bolts is selected and the diameter recorded. If the resulting sample mean is less than 0.49 inches or greater than 0.51 inches, the process is shut down for adjustment. The standard deviation for the diameter of a single bolt is 0.03 inches. What is the probability that the sample mean fall outside the 0.49 to 0.51 inch interval? 0.005 z = -2 z = 2 0.046 normalcdf(-2,2,0,1) 0.046 normalcdf(-99,-2,0,1) normalcdf(2,99,0,1)

39. Probability #11 A patient with a failing heart has been told that the probability of finding a suitable transplant is.36, the conditional probability of surviving given that a transplant operation is performed is.72, and the conditional probability of surviving given that a transplant operation is not performed is.22. Find:

40. Probability #11 A patient with a failing heart has been told that the probability of finding a suitable transplant is.36, the conditional probability of surviving given that a transplant operation is performed is.72, and the conditional probability of surviving given that a transplant operation is not performed is.22. Find: Transplant.36.64 No Transplant Survive.72.28 Die Survive.22.78 Die

41. Probability #11 A patient with a failing heart has been told that the probability of finding a suitable transplant is.36, the conditional probability of surviving given that a transplant operation is performed is.72, and the conditional probability of surviving given that a transplant operation is not performed is.22. Find: A) The probability the patient survives P(Transplant & Live) = P(No Transplant & Live) =.259.141 0.4

42. Probability #11 A patient with a failing heart has been told that the probability of finding a suitable transplant is.36, the conditional probability of surviving given that a transplant operation is performed is.72, and the conditional probability of surviving given that a transplant operation is not performed is.22. Find: B)The probability a suitable transplant is found and the patient survives P(Transplant & Live) = P(No Transplant & Live) =.259.141 0.259

43. Probability #11 A patient with a failing heart has been told that the probability of finding a suitable transplant is.36, the conditional probability of surviving given that a transplant operation is performed is.72, and the conditional probability of surviving given that a transplant operation is not performed is.22. Find: C)The probability that a suitable transplant was found given the patient survived P(Transplant & Live) = P(No Transplant & Live) =.259.141.4 P(A | B) =