General Rules of Probability BPS 7e Chapter 13 © 2015 W. H. Freeman and Company

Independence A bag contains 10 marbles, 3 of which are red. A marble is drawn at random from the bag. The marble is replaced, and then another marble is drawn from the bag. Consider the events A = the first marble is red and B = the second marble is red. Are A and B independent? yes no impossible to say

Independence (answer) A bag contains 10 marbles, 3 of which are red. A marble is drawn at random from the bag. The marble is replaced, and then another marble is drawn from the bag. Consider the events A = the first marble is red and B = the second marble is red. Are A and B independent? yes no impossible to say

Multiplication Rule Three students work independently on a homework problem. The probability that the first student solves the problem is 0.95. The probability that the second student solves the problem is 0.85. The probability that the third student solves the problem is 0.80. What is the probability that all three solve the problem? 0.95 + 0.85 + 0.80 (0.95) (0.85) (0.80) 1 – 0.95 – 0.85 – 0.80 1 – (0.95) (0.85) (0.80) 0.80

Multiplication Rule (answer) Three students work independently on a homework problem. The probability that the first student solves the problem is 0.95. The probability that the second student solves the problem is 0.85. The probability that the third student solves the problem is 0.80. What is the probability that all three solve the problem? 0.95 + 0.85 + 0.80 (0.95) (0.85) (0.80) 1 – 0.95 – 0.85 – 0.80 1 – (0.95) (0.85) (0.80) 0.80

Multiplication Rule Three students work independently on a homework problem. The probability that the first student solves the problem is 0.95. The probability that the second student solves the problem is 0.85. The probability that the third student solves the problem is 0.80. What is the probability that the first student solves the problem and the other two students do not? 0.95 + 0.15 + 0.20 (0.95) (0.15) (0.20) 0.95 – 0.15 – 0.20 0.95 – 0.85 – 0.80 0.95

Multiplication Rule (answer) Three students work independently on a homework problem. The probability that the first student solves the problem is 0.95. The probability that the second student solves the problem is 0.85. The probability that the third student solves the problem is 0.80. What is the probability that the first student solves the problem and the other two students do not? 0.95 + 0.15 + 0.20 (0.95) (0.15) (0.20) 0.95 – 0.15 – 0.20 0.95 – 0.85 – 0.80 0.95

Multiplication Rule Three students work independently on a homework problem. The probability that the first student solves the problem is 0.95. The probability that the second student solves the problem is 0.85. The probability that the third student solves the problem is 0.80. What is the probability that none of the three students solves the problem? 1 – 0.95 – 0.85 – 0.80 0.05 + 0.15 + 0.20 1 – (0.95) (0.85) (0.80) (0.05) (0.15) (0.20)

Multiplication Rule (answer) Three students work independently on a homework problem. The probability that the first student solves the problem is 0.95. The probability that the second student solves the problem is 0.85. The probability that the third student solves the problem is 0.80. What is the probability that none of the three students solves the problem? 1 – 0.95 – 0.85 – 0.80 0.05 + 0.15 + 0.20 1 – (0.95) (0.85) (0.80) (0.05) (0.15) (0.20)

Multiplication Rule Three students work independently on a homework problem. The probability that the first student solves the problem is 0.95. The probability that the second student solves the problem is 0.85. The probability that the third student solves the problem is 0.80. What is the probability that at least one student solves the problem correctly? 0.95 + 0.85 + 0.80 (0.95) (0.85) (0.80) 1 – (0.05) (0.15) (0.20) 0.95 0.80

Multiplication Rule (answer) Three students work independently on a homework problem. The probability that the first student solves the problem is 0.95. The probability that the second student solves the problem is 0.85. The probability that the third student solves the problem is 0.80. What is the probability that at least one student solves the problem correctly? 0.95 + 0.85 + 0.80 (0.95) (0.85) (0.80) 1 – (0.05) (0.15) (0.20) 0.95 0.80

Multiplication Rule A candy company manufactures bags of colorful candies. The company reports that it makes 10% each of green and red candies and 20% each of yellow, blue, and orange candies. The rest of the candies are brown. If you pick four candies in a row, what is the probability that the fourth pick is the first green candy? 0.8311 0.0687 0.0932 0.0729

Multiplication Rule (answer) A candy company manufactures bags of colorful candies. The company reports that it makes 10% each of green and red candies and 20% each of yellow, blue, and orange candies. The rest of the candies are brown. If you pick four candies in a row, what is the probability that the fourth pick is the first green candy? 0.8311 0.0687 0.0932 0.0729 P(fourth pick that is first green) = P(not green) × P(not green) × P(not green) × P(green) = (0.9) × (0.9) × (0.9) × (0.1) = 0.0729

Multiplication Rule An instant lottery game gives you probability 0.03 of winning on any one play. Plays are independent of each other. If you are to play four times in a row, the probability that you do NOT win in any four consecutive plays is about: 0.0009. 0.12. 0.885. 0.912.

Multiplication Rule (answer) An instant lottery game gives you probability 0.03 of winning on any one play. Plays are independent of each other. If you are to play four times in a row, the probability that you do NOT win in any four consecutive plays is about” 0.0009. 0.12. 0.885. (0.97 * 0.97 * 0.97 * 0.97) 0.912.

General Addition Rule A survey of local car dealers revealed that 64% of all cars sold last month had CD players, 28% had alarm systems, and 22% had both CD players and alarm systems. What is the probability that a car has either a CD player or an alarm system? 0.64 + 0.22 – 0.28 (0.64) (0.28) 0.64 + 0.28 – 0.22 0.64 + 0.28 – (0.22) (0.64)

General Addition Rule (answer) A survey of local car dealers revealed that 64% of all cars sold last month had CD players, 28% had alarm systems, and 22% had both CD players and alarm systems. What is the probability that a car has either a CD player or an alarm system? 0.64 + 0.22 – 0.28 (0.64) (0.28) 0.64 + 0.28 – 0.22 0.64 + 0.28 – (0.22) (0.64)

General Addition Rule Three students work independently on a homework problem. The probability that the first student solves the problem is 0.95. The probability that the second student solves the problem is 0.85. The probability that the third student solves the problem is 0.80. What is the probability that either the first or the second student solves the problem? 0.95 + 0.85 – 0.80 (0.95) (0.85) 0.95 + 0.85 – (0.95) (0.85) 0.95 + 0.85

General Addition Rule (answer) Three students work independently on a homework problem. The probability that the first student solves the problem is 0.95. The probability that the second student solves the problem is 0.85. The probability that the third student solves the problem is 0.80. What is the probability that either the first or the second student solves the problem? 0.95 + 0.85 - 0.80 (0.95) (0.85) 0.95 + 0.85 – (0.95) (0.85) 0.95 + 0.85

General Addition Rule A student takes English and American History this semester. The probability that a student passes English is 0.50. The probability that a student passes American History is 0.40. The probability that a student passes both English and American History is 0.30. What is the probability that the student passes either English or American History or both? 0.50 + 0.40 0.50 + 0.40 – 0.30 0.30 – 0.50 – 0.40 0.50 + 0.40 + 0.30 (0.50) (0.40) – 0.30

General Addition Rule (answer) A student takes English and American History this semester. The probability that a student passes English is 0.50. The probability that a student passes American History is 0.40. The probability that a student passes both English and American History is 0.30. What is the probability that the student passes either English or American History or both? 0.50 + 0.40 0.50 + 0.40 – 0.30 0.30 – 0.50 – 0.40 0.50 + 0.40 + 0.30 (0.50) (0.40) – 0.30

General Addition Rule A candy company manufactures bags of colorful candies. The company reports that it makes 10% each of green and red candies and 20% each of yellow, blue, and orange candies. The rest of the candies are brown. If you pick a candy at random, what is the probability that it is EITHER red or yellow? 0.10 0.20 0.30 0.40

General Addition Rule (answer) A candy company manufactures bags of colorful candies. The company reports that it makes 10% each of green and red candies and 20% each of yellow, blue, and orange candies. The rest of the candies are brown. If you pick a candy at random, what is the probability that it is EITHER red or yellow? 0.10 0.20 0.30 (P(red) + P(yellow) = 0.10 + 0.20 = 0.30) 0.40

General Multiplication Rule Within the United States, approximately 11.25% of the population is left-handed. Of the males, 12.6% are left-handed, compared with only 9.9% of the females. The probability of selecting a male is the same as that of selecting a female. If a person is selected at random, what is the probability that the person is a left-handed male? 0.126 / 0.50 (0.126) (0.1125) (0.50) (0.126) (0.50)

General Multiplication Rule (answer) Within the United States, approximately 11.25% of the population is left-handed. Of the males, 12.6% are left-handed, compared twith only 9.9% of the females. The probability of selecting a male is the same as that of selecting a female. If a person is selected at random, what is the probability that the person is a left-handed male? 0.126 / 0.50 (0.126) (0.1125) (0.50) (0.126) (0.50)

General Multiplication Rule Within the United States, approximately 11.25% of the population is left-handed. Of the males, 12.6% are left-handed, compared with only 9.9% of the females. The probability of selecting a male is the same as that of selecting a female. If a person is selected at random, what is the probability that the person is a right-handed female? (0.099) (0.50) (0.099) (0.901) (0.50) 0.901

General Multiplication Rule (answer) Within the United States, approximately 11.25% of the population is left-handed. Of the males, 12.6% are left-handed, compared with only 9.9% of the females. The probability of selecting a male is the same as that of selecting a female. If a person is selected at random, what is the probability that the person is a right-handed female? (0.099) (0.50) (0.099) (0.901) (0.50) 0.901

Conditional Probability The probability that a randomly selected fourth-grade student is proficient in math is 0.85. The probability that such a student is proficient in reading is 0.78. The probability that such a student is proficient in both math and reading is 0.65. What is the conditional probability that such a student is proficient in reading given that they are proficient in math? 0.85 / 0.65 0.65 / 0.85 0.85 / 0.78 0.65 / 0.78

Conditional Probability (answer) The probability that a randomly selected fourth-grade student is proficient in math is 0.85. The probability that such a student is proficient in reading is 0.78. The probability that such a student is proficient in both math and reading is 0.65. What is the conditional probability that such a student is proficient in reading given that they are proficient in math? 0.85 / 0.65 0.65 / 0.85 0.85 / 0.78 0.65 / 0.78

Conditional Probability A student takes English and American History this semester. The probability that a student passes English is 0.50. The probability that a student passes American History is 0.40. The probability that a student passes both English and American History is 0.30. Are passing English and passing American History independent events? yes, because the classes are taught by different teachers yes, because the classes use different skills no, because (0.50) (0.40)  0.30 no, because (0.50)  (0.40) (0.30)

Conditional Probability (answer) A student takes English and American History this semester. The probability that a student passes English is 0.50. The probability that a student passes American History is 0.40. The probability that a student passes both English and American History is 0.30. Are passing English and passing American History independent events? yes, because the classes are taught by different teachers yes, because the classes use different skills no, because (0.50) (0.40)  0.30 no, because (0.50)  (0.40) (0.30)

Conditional Probability At a certain university, 47.0% of the students are female. Also, 8.5% of the students are married females. If a student is selected at random, what is the probability that the student is married given that the student was female? 0.085 / 0.47 0.47 / 0.085 (0.085) (0.47) 0.085 0.47

Conditional Probability (answer) At a certain university, 47.0% of the students are female. Also, 8.5% of the students are married females. If a student is selected at random, what is the probability that the student is married given that the student was female? 0.085 / 0.47 0.47 / 0.085 (0.085) (0.47) 0.085 0.47

Independence Again At a local high school, the probability that a student is absent on any particular day is about 0.08. However, the probability that a student is absent given that it is Friday is 0.20. Are the events “absent” and “Friday” independent? no, because knowing it is Friday changes the probability of a student being absent yes, because being absent has nothing to do with what day of the week it is cannot be determined

Independence Again (answer) At a local high school, the probability that a student is absent on any particular day is about 0.08. However, the probability that a student is absent given that it is Friday is 0.20. Are the events “absent” and “Friday” independent? no, because knowing it is Friday changes the probability of a student being absent yes, because being absent has nothing to do with what day of the week it is cannot be determined

Tree Diagrams Kicking a field goal is the result of three processes: snapping, holding, and kicking. The probability that a specific kicker makes a field goal from 35 yards given that the hold is good is 0.80. The probability that the hold is good given that the snap is good is 0.95. The probability that the snap is good is 0.99. What is the probability that the snap is good, the hold is good, and the kick is good? 0.80 (0.80) (0.95) + (.80) (0.05) (0.80) (0.95) (0.99) (0.80) (0.95) (0.99) + (0.80) (0.95) (0.01)

Tree Diagrams (answer) Kicking a field goal is the result of three processes: snapping, holding, and kicking. The probability that a specific kicker makes a field goal from 35 yards given that the hold is good is 0.80. The probability that the hold is good given that the snap is good is 0.95. The probability that the snap is good is 0.99. What is the probability that the snap is good, the hold is good, and the kick is good? 0.80 (0.80) (0.95) + (.80) (0.05) (0.80) (0.95) (0.99) (0.80) (0.95) (0.99) + (0.80) (0.95) (0.01)