STATPRO Exercises with Solutions. Problem Set A: Basic Probability

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1 Problem Set A: Basic Probability 1. A tea taster is required to taste and rank three varieties of tea namely Tea A, B and C; according to the tasters preference. (ranking the teas from the best choice until the tasters least choice) a) Define the experiment Ranking the three varieties of tea from the tasters preference (e.g. first, second, last choice) b) List the sample points in the sample space. (Tasters order of preference: first choice, second choice and last choice) A B C B C A A C B C A B B A C C B A c) If the taster had no ability to distinguish a difference in the taste between teas, what is the probability that the taster will rank Tea A as best? Let A be the event that the taster ranked Tea A as the best. P(A) = n N = 2 6 = 1 3 d) Using the same condition in c) what is the probability that the taster will rank Tea A as least desirable? Let B be the event that the taster ranked Tea A as the least desirable. P(B) = n N = 2 6 = Two dice are tossed. What is the probability that the sum of the numbers showing on the dice is equal to 9? Prepared by Dr. Francis Joseph H. Campeña 1

2 Let C be the event that the sum of the numbers showing on the dice is equal to 9. The following shows the possible outcomes of event C when a pair of dice is tossed. P(C) = n N = 4 36 = In how many different ways can a true-false test consisting of questions be answered? We use the fundamental principle of counting in finding the number of ways of answering the test. Two factorial ways of answering question number 1, two factorial ways of answering question number 2 and so on (2!)(2!)(2!)(2!)(2!)(2!)(2!)(2!)(2!)(2!) = 2 = 1, How many distinct permutations can be made to the letters of the word columns. Since all the letters in the word columns are all distinct we can just use the formula n! to determine the number of distinct permutations of the letters in the word columns. n! = (7)! = 5, A college plays 12 football games during a season. In how many ways can the team end the season with 7 wins, 3 losses and 2 ties? 12C 7 5 C 3 2 C 2 Selecting a game where the team won Selecting a game where the team lost Selecting a game where the team tied to another team 6. From a group of 4 men and 5 women, how many committees of size 3 are possible a. With no restrictions? 9! n = 9 C 3 = (9 3)! 3! = 9! 6! 3! = 84 Prepared by Dr. Francis Joseph H. Campeña 2

3 b. With 1 man and 2 women? n = 4 C 1 5 C 2 = 4 = 40 Selecting a man Selecting a woman c. With 2 men and 1 woman if a certain man must be on the committee? If a certain man must be in the committee, we can now only chose one man and a woman to form the committee. n = 3 C 1 5 C 1 = 3 5 = 15 Selecting a man Selecting a woman Problem Set B: Laws on Probability 1. A certain genetic characteristic occurs in mice with probability equal to 0.2. If two mice are randomly selected from a large number of unrelated litters, what is the probability that both mice possess the genetic characteristic? Suppose events A and B are events of selecting the first and second mouse having a certain genetic characteristic. Since the mice are randomly selected, and the probability that the second mouse has the genetic characteristics does not depend on the probability that the first mouse also has the genetic characteristic we have the following: P(A B) = P(A) P(B) = (0.2)(0.2) = A lineup of men is conducted to test the ability of a witness to identify three burglary suspects. Suppose that the three burglary suspects who committed the crime are in the lineup. If the witness is actually unable to identify the suspects but feels compelled to make a choice, a) What is the probability that the three guilty men are selected by chance? Let A be the event that the three guilty men are selected by the witness. P(A) = n = 3 C 3 = 1 N C b) What is the probability that the witness selects three innocent men? Let B be the event that the three innocent men are selected by the witness. P(B) = n = 7 C 3 = 35 = 7 N C Prepared by Dr. Francis Joseph H. Campeña 3

4 3. If 3 books are picked at random from a shelf containing 5 novels, 3 books of poems and a dictionary, what is the probability that a) The dictionary is selected? Let B be the event that the dictionary is selected. P(A) = n = 8 C 2 1 C 1 = 28 = 1 N 9C b) 2 novels and 1 book of poems are selected? Let B be the event that the dictionary is selected. P(A) = n = 5 C 2 3 C 1 = 30 = 5 N 9C In a college graduating class of 0 students, 54 studied mathematics, 69 studied history and 35 studied both mathematics and history. If one of these students is selected at random, find the probability that Let A be the event that the student selected studied mathematics B be the event that the student selected studied history P(A) = 54 0 P(B) = 69 0 P(A B) = 35 0 a) The student takes mathematics or history P(A B) = P(A) + P(B) P(A B) = = 88 0 b) The student does not take either of these subjects The event (A B) c is the event where the student does not take either the math or history subjects. P(A B) + P((A B) c ) = P((A B)c ) = 1 P((A B) c ) = = 12 0 c) The student takes history but not mathematics The event A c B is the event where student takes history but not mathematics - = Prepared by Dr. Francis Joseph H. Campeña 4

5 Thus we have the following: P(A c B ) = P(A) P(A B) = = The probability that a married man watches a telenovelas is 0.4 and the probability that a married woman watches the show is 0.8. The probability that a man watches the show given that his wife does is Find the probability that Let M be the event that a married man watches telenovelas is selected W be the event that a married woman watches telenovelas is selected P(M) = 0.4 P(W) = 0.8 P(M W) = 0.85 a) A married couple watches the show P(M W) = P(M W) P(W) = (0.085)(0.8) = b) A wife watched the show given that her husband does P(M W) P(W M) = = P(M) 0.4 = 0.17 c) At least 1 person of a married couple will watch the show P(M W) = P(M) + P(W) P(M W) = = Problem Set C: Normal Distribution 1. Given a normal distribution with μ = 40 and σ 2 = 0 σ =. Find a. P(X < 32) P(X < 32) = P (z < ) = P (z < 8 ) = P(z < 0.8) = b. P(X > 27) P(X > 27) = 1 P (z < ) = 1 P(z < 1.3) = = c. P(42 < X < 51) P(42 < x < 51) = P(z < 51) P(z < 42) = P (z < ) P (z < = P (z < 11 ) -P (z < 2 ) = P(z < 1.1)-P(z < 0.2) = = ) Prepared by Dr. Francis Joseph H. Campeña 5

6 2. Given a normal distribution with μ = 200 and σ 2 = 0 σ =. Find a. P(X < 214) P(X < 214) = P (z < ) = P(z < 1.4) = b. P(X > 179) P(X > 179) = 1 P(X < 179) = 1 P (z < ) = 1 P(z < 2.1) = = c. P(188 < X < 206) P(188 < X < 206) = P(X < 206) P(X < 188) = P (z < ) P (z < = P(z < 0.6)-P(z < -1.2) ) 3. Adult female have forearm length that are normally distributed with mean 17.5 in and standard deviation of 0.75 in. Remark: Let X be a random variable for the forearm length of an adult female. a. Find the probability that a female s forearm length is between 16 in and 18 in. P(16 < x < 18) = P(x < 18) P(x < 16) = P (z < ) P (x < 0.75 = P(z < 0.67)-P(x < -2.0) = = b. Find the that a female s forearm length is greater than 18.5 P(x > 18.5) = 1 P(x < 18.5) = 1 P (z < ) 0.75 = 1-P(z < 1.33) = = ) A chemical process requires a ph between 5.85 and 7.4. The ph of the process is a normal random variable with mean 6.0 and standard deviation 0.9. Assume that the process must be shut down if the ph will fall outside the acceptable range of 5.85 to Prepared by Dr. Francis Joseph H. Campeña 6

7 Let X be a random variable that denotes the ph level of the chemical process. μ = 6.0 and σ = 0.9 a. What is the probability that the process will not be shut down? Since the threshold of the ph level is from We want to know the probability that the ph level is within this values. P(5.85 < x < 7.40) = P(x < 7.40) P(x < 5.85) = P (z < ) P (x < ) = P(z < 1.56) P(x < 0.17) = = b. What is the probability that the process will be shut down? We let A be the event that the ph level of the process is between 5.58 and 7.40 then A c denotes that the ph level is beyond the threshold of the chemical process and from 4a) P(A) = If we want to get the probability that the process will be shut down we have to get the probability of A c. Thus, P(A c ) = 1 P(A) = = Heights of adult males are normally distributed with mean 65 in and standard deviation of 2.5 in. A male is selected at random. Let X be a random variable that denotes the heights of adult males. μ = 65 and σ = 2.5 a. Find the probability that a male selected have a height greater than 67. P(x > 67) = 1 P(x < 67) = 1 P (z < 2.5 ) = 1 P(z < 0.8) = = b. Find the probability that a male selected have a height less than P(x < 60) = P (z < 2.5 ) = P(z < 2) = Assume that the age at onset of disease X is normally distributed with mean of 50 years and a standard deviation of 15 years. What is the probability that an individual is afflicted with disease X developed it before age 35? Prepared by Dr. Francis Joseph H. Campeña 7

8 Let X be a random variable that denotes the age when an individual is afflicted with the disease X The following data are given: μ = 50 and σ = P(x < 35) = P (z < ) = P(z < 1) = If adult male cholesterol is normally distributed with μ = 200 and σ = 30, what is the probability of selecting a male whose cholesterol is Let X be a random variable that denotes the cholesterol level of an adult male The following data are given: μ = 200 and σ = 30 a. Less than P(x < 165) = P (z < ) 30 = P(z < 1.17) = 0.12 b. Greater than 165 P(x > 165) = 1 P(x < 165) = = c. Between 165 and 220 P(165 < x < 220) = P(x < 220) P(x < 165) = P (z < ) P (z < 30 = P(z < 0.67) P(z < 1.17) = = d. Greater than 220 P(x > 220) = 1 P(x < 220) = 1 P (z < ) 30 = 1 P(z < 0.67) = = ) 30 Prepared by Dr. Francis Joseph H. Campeña 8

Chapter 2 PROBABILITY SAMPLE SPACE

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