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Math I Probability Experiments Probability experiment • An action, or trial, through which specific results (counts, measurements, or responses) are obtained. Outcome • The result of a single trial in a probability experiment. Sample Space • The set of all possible outcomes of a probability experiment. Event • Consists of one or more outcomes and is a subset of the sample space. Larson/Farber 4th ed 3 Probability Experiments • Probability experiment: Roll a die • Outcome: {3} • Sample space: {1, 2, 3, 4, 5, 6} • Event: {Die is even}={2, 4, 6} Larson/Farber 4th ed 4 Example: Identifying the Sample Space A probability experiment consists of tossing a coin and then rolling a six-sided die. Describe the sample space. Solution: There are two possible outcomes when tossing a coin: a head (H) or a tail (T). For each of these, there are six possible outcomes when rolling a die: 1, 2, 3, 4, 5, or 6. One way to list outcomes for actions occurring in a sequence is to use a tree diagram. Larson/Farber 4th ed 5 Solution: Identifying the Sample Space Tree diagram: H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6 The sample space has 12 outcomes: {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6} Larson/Farber 4th ed 6 Simple Events Simple event • An event that consists of a single outcome. e.g. “Tossing heads and rolling a 3” {H3} • An event that consists of more than one outcome is not a simple event. e.g. “Tossing heads and rolling an even number” {H2, H4, H6} Larson/Farber 4th ed 7 Example: Identifying Simple Events Determine whether the event is simple or not. • You roll a six-sided die. Event B is rolling at least a 4. Solution: Not simple (event B has three outcomes: rolling a 4, a 5, or a 6) Larson/Farber 4th ed 8 Probability is the chance that something will happen. Probability is most often expressed as a fraction, a decimal, a percent, or can also be written out in words. To determine the probability… P(event) = number of true outcomes total number of equally likely outcomes Independent: Two events are independent if the occurrence of one has no effect on the occurrence of the other… Example 1: (probability of two events) What is the probability of drawing a king and then an ace from a standard 52 card deck with replacement? 4 4 16 P(King, Ace) = 52 52 2704 Example 2: What is the probability of flipping heads on a coin three times in a row? 1 1 1 1 P(H, H, H) = 2 2 2 8 Example 3: A die is rolled twice. What’s the probability of rolling a 2 and then an even number? 1 1 1 Solution: 6 2 12 Example 4: You spin the spinner 3 times. What is the probability of spinning a 4, a 3 and then a 1? Solution: 1 2 3 6 3 8 8 8 512 256 Complementary Events Complement of event E • The set of all outcomes in a sample space that are not included in event E. • Denoted E ′ (E prime) • P(E ′) + P(E) = 1 • P(E) = 1 – P(E ′) E′ • P(E ′) = 1 – P(E) E Larson/Farber 4th ed 13 Let’s try a few Complement Spinner • In the spinner at the right: • P’(yellow) • P’(not blue) • P’(red or yellow) • P’(white) • P’(odd number) Dependent: Two events such that the occurrence of one affects the occurrence of the other. P(A and B) = P(A) P(B|A) **P(B|A) = means the probability of B given that event A has already occurred. Dependent: WITHOUT REPLACEMENT Independent: WITH REPLACEMENT Independent and Dependent Events Independent events • The occurrence of one of the events does not affect the probability of the occurrence of the other event • P(B | A) = P(B) or P(A | B) = P(A) • Events that are not independent are dependent Larson/Farber 4th ed 18 Example: Independent and Dependent Events Decide whether the events are independent or dependent. 1. Selecting a king from a standard deck (A), not replacing it, and then selecting a queen from the deck (B). Solution: 4 P( B | A) P(2nd card is a Queen |1st card is a King ) 51 4 P( B) P(Queen) 52 Dependent (the occurrence of A changes the probability of the occurrence of B) Larson/Farber 4th ed 19 Example: Independent and Dependent Events Decide whether the events are independent or dependent. 2. Tossing a coin and getting a head (A), and then rolling a six-sided die and obtaining a 6 (B). Solution: 1 P( B | A) P(rolling a 6 | head on coin) 6 1 P( B) P(rolling a 6) 6 Independent (the occurrence of A does not change the probability of the occurrence of B) Larson/Farber 4th ed 20 You Spin A Spinner Twice ◦ Independent You choose a pair of shoes and a pair of socks ◦ Independent You buy two different CD’s ◦ Dependent You roll a number cube twice so that6 the sum is 8 ◦ Dependent You spin a spinner and flip a coin ◦ Independent You pick a marble and pick again without replacing. ◦ Dependent Example: Using the Multiplication Rule Two cards are selected, without replacing the first card, from a standard deck. Find the probability of selecting a king and then selecting a queen. Solution: Because the first card is not replaced, the events are dependent. P( K and Q) P ( K ) P (Q | K ) 4 4 52 51 16 0.006 2652 Larson/Farber 4th ed 23 Example: Using the Multiplication Rule A coin is tossed and a die is rolled. Find the probability of getting a head and then rolling a 6. Solution: The outcome of the coin does not affect the probability of rolling a 6 on the die. These two events are independent. P ( H and 6) P ( H ) P (6) 1 1 2 6 1 0.083 12 Larson/Farber 4th ed 24 Example: Using the Multiplication Rule The probability that a particular knee surgery is successful is 0.85. Find the probability that three knee surgeries are successful. Solution: The probability that each knee surgery is successful is 0.85. The chance for success for one surgery is independent of the chances for the other surgeries. P(3 surgeries are successful) = (0.85)(0.85)(0.85) ≈ 0.614 Larson/Farber 4th ed 25 Example: Using the Multiplication Rule Find the probability that none of the three knee surgeries is successful. Solution: Because the probability of success for one surgery is 0.85. The probability of failure for one surgery is 1 – 0.85 = 0.15 P(none of the 3 surgeries is successful) = (0.15)(0.15)(0.15) ≈ 0.003 Larson/Farber 4th ed 26 Example: Using the Multiplication Rule Find the probability that at least one of the three knee surgeries is successful. Solution: “At least one” means one or more. The complement to the event “at least one successful” is the event “none are successful.” Using the complement rule P(at least 1 is successful) = 1 – P(none are successful) ≈ 1 – 0.003 = 0.997 Larson/Farber 4th ed 27 Example: Using the Multiplication Rule to Find Probabilities More than 15,000 U.S. medical school seniors applied to residency programs in 2007. Of those, 93% were matched to a residency position. Seventy-four percent of the seniors matched to a residency position were matched to one of their top two choices. Medical students electronically rank the residency programs in their order of preference and program directors across the United States do the same. The term “match” refers to the process where a student’s preference list and a program director’s preference list overlap, resulting in the placement of the student for a residency position. (Source: National Resident Matching Program) (continued) Larson/Farber 4th ed 28 Example: Using the Multiplication Rule to Find Probabilities 1. Find the probability that a randomly selected senior was matched a residency position and it was one of the senior’s top two choices. Solution: A = {matched to residency position} B = {matched to one of two top choices} P(A) = 0.93 and P(B | A) = 0.74 P(A and B) = P(A)∙P(B | A) = (0.93)(0.74) ≈ 0.688 dependent events Larson/Farber 4th ed 29 Example: Using the Multiplication Rule to Find Probabilities 2. Find the probability that a randomly selected senior that was matched to a residency position did not get matched with one of the senior’s top two choices. Solution: Use the complement: P(B′ | A) = 1 – P(B | A) = 1 – 0.74 = 0.26 Larson/Farber 4th ed 30 Example 1: What is the probability of drawing a King and then an Ace without replacement? 4 4 16 P(King, Ace) = 52 51 2652 Example 2: You randomly select two marbles from a bag that contains 14 green, 7 blue, and 9 red marbles. What is the probability that the first marble is blue and the second marble is not blue if you do not replace the first marble? Solution: 7 23 161 30 29 870 Example 3: Your teacher passes around a basket with 6 red erasers, 9 blue erasers, and 7 green erasers. If you and your two neighbors are the first to randomly select an eraser, what is the probability that all three of you select green erasers? Solution: P(A) and P(B|A) and P(C|A and B) 7 6 5 210 22 21 20 9240 The table shows the number of males and females with certain hair colors. Find … A) the probability that a listed person has red hair B) the probability that a female has red hair Brown Blonde Red Black Other hair hair hair hair Male 42 11 3 17 27 Female 47 16 13 9 15 P(red hair) = # of people with red hair 16 2 total # of people 200 25 P(red hair | female) = # of red hair females 13 total # of females 100 When you consider the outcomes for either of two events A and B, you form the union of A and B. When you consider only the outcomes shared by both A and B, you form the intersection of A and B. When the sets of A and B have nothing in common (no intersection) then they are considered mutually exclusive events. Mutually exclusive Two events A and B cannot occur at the same time A B A B A and B are A and B are not mutually mutually exclusive exclusive 40 Larson/Farber 4th ed Decide if the events are mutually exclusive. Event A: Roll a 3 on a die. Event B: Roll a 4 on a die. Solution: Mutually exclusive (The first event has one outcome, a 3. The second event also has one outcome, a 4. These outcomes cannot occur at the same time.) 41 Larson/Farber 4th ed Decide if the events are mutually exclusive. Event A: Randomly select a male student. Event B: Randomly select a nursing major. Solution: Not mutually exclusive (The student can be a male nursing major.) 42 Larson/Farber 4th ed If A and B are mutually exclusive events (one event does not have anything in common with the other), then… P(A or B) = P(A) + P(B) A die is rolled one time. What is the probability of rolling a 2 or a 6? Solution: 1 1 1 2 P(A or B) = P(A) + P(B) = 6 6 6 3 A card is randomly selected out of a standard deck of 52 cards. What is the probability that it is a 2 or a king? P(A or B) = P(A) + P(B) = 2 4 52 52 4 8 52 13 If A and B are not mutually exclusive, then there are some outcomes in common. Therefore, the intersection of A and B are counted twice when P(A) and P(B) are added. So, P(A and B) must be subtracted once from the sum… P(A or B) = P(A) + P(B) – P(A and B) A die is rolled one time. What is the probability of rolling an odd number or a prime number? 3 Odd = 1, 3, 5 P(A) = 6 Prime = 2, 3, 5 P(B) = 3 6 2 Odd and prime = 3, 5 P(A and B) = 6 A card is randomly selected from standard deck of 52 cards. What is the probability that it is a red card or a king? Red cards = 26 = 26 52 Kings = 4 = 4 52 2 Red Kings = 2 = 52 The probability that it will rain today is 40%. The probability that is will rain tomorrow is 30%. The probability that it will rain both days is 20%. Find the probability that it will rain either today OR tomorrow. Solution: P(A) + P(B) – P(A and B) P(today) + P(tomorrow) – P(today and tomorrow) = 40% 30% 20% 50% The Addition Rule Addition rule for the probability of A or B • The probability that events A or B will occur is P(A or B) = P(A) + P(B) – P(A and B) • For mutually exclusive events A and B, the rule can be simplified to P(A or B) = P(A) + P(B) Can be extended to any number of mutually exclusive events Larson/Farber 4th ed 50 Example: Using the Addition Rule You select a card from a standard deck. Find the probability that the card is a 4 or an ace. Solution: The events are mutually exclusive (if the card is a 4, it cannot be an ace) Deck of 52 Cards P (4 or ace) P (4) P( ace) 4♣ 4 4 4♥ 4♠ A♣ 52 52 4♦ A♠ A♥ 8 A♦ 0.154 44 other cards 52 Larson/Farber 4th ed 51 Example: Using the Addition Rule You roll a die. Find the probability of rolling a number less than 3 or rolling an odd number. Solution: The events are not mutually exclusive (1 is an outcome of both events) Roll a Die 4 6 Odd Less than 3 1 three 5 2 Larson/Farber 4th ed 52 Solution: Using the Addition Rule Roll a Die 4 6 Odd Less than 3 1 three 5 2 P(less than 3 or odd ) P(less than 3) P(odd ) P(less than 3 and odd ) 2 3 1 4 0.667 6 6 6 6 Larson/Farber 4th ed 53 Example: Using the Addition Rule The frequency distribution shows Sales volume ($) Months the volume of sales (in dollars) 0–24,999 3 and the number of months a sales 25,000–49,999 5 representative reached each sales 50,000–74,999 6 level during the past three years. 75,000–99,999 7 If this sales pattern continues, 100,000–124,999 9 what is the probability that the 125,000–149,999 2 sales representative will sell 150,000–174,999 3 between $75,000 and $124,999 175,000–199,999 1 next month? Larson/Farber 4th ed 54 Solution: Using the Addition Rule • A = monthly sales between Sales volume ($) Months $75,000 and $99,999 0–24,999 3 • B = monthly sales between 25,000–49,999 5 $100,000 and $124,999 50,000–74,999 6 • A and B are mutually exclusive 75,000–99,999 7 100,000–124,999 9 P ( A or B ) P ( A) P ( B ) 125,000–149,999 2 7 9 150,000–174,999 3 36 36 175,000–199,999 1 16 0.444 36 Larson/Farber 4th ed 55 Example: Using the Addition Rule A blood bank catalogs the types of blood given by donors during the last five days. A donor is selected at random. Find the probability the donor has type O or type A blood. Type O Type A Type B Type AB Total Rh-Positive 156 139 37 12 344 Rh-Negative 28 25 8 4 65 Total 184 164 45 16 409 Larson/Farber 4th ed 56 Solution: Using the Addition Rule The events are mutually exclusive (a donor cannot have type O blood and type A blood) Type O Type A Type B Type AB Total Rh-Positive 156 139 37 12 344 Rh-Negative 28 25 8 4 65 Total 184 164 45 16 409 P (type O or type A) P (type O ) P (type A) 184 164 409 409 348 0.851 409 Larson/Farber 4th ed 57 Example: Using the Addition Rule Find the probability the donor has type B or is Rh- negative. Type O Type A Type B Type AB Total Rh-Positive 156 139 37 12 344 Rh-Negative 28 25 8 4 65 Total 184 164 45 16 409 Solution: The events are not mutually exclusive (a donor can have type B blood and be Rh-negative) Larson/Farber 4th ed 58 Solution: Using the Addition Rule Type O Type A Type B Type AB Total Rh-Positive 156 139 37 12 344 Rh-Negative 28 25 8 4 65 Total 184 164 45 16 409 P(type B or Rh neg ) P(type B) P( Rh neg ) P(type B and Rh neg ) 45 65 8 102 0.249 409 409 409 409 Larson/Farber 4th ed 59 Formula Means the probability of B given that A has already occurred equals ◦ The probability of A and B occurring Divided by the probability of A Example 1: ◦ Gus rolled a six-sided cube, with faces numbered 1 to 6, twice. If you know that he rolled a 2 on his first roll, what is the probability that he rolled two even numbers? Find the probability of getting two even numbers given that he rolled a two Probability of A and B is 1/6 times ½ which equals 1/12 Divide that by 1/6 = 1/2 Example 2: ◦ Antoine will toss a fair coin three times. What is the probability that all 3 tosses will result in heads if you know that his first toss resulted in heads? Find the probability of getting all three heads given that the first one is heads Probability of getting all three heads is ½ times ½ times ½ which is 1/8 Divide that by 1/2 = 1/4 Example 3: A bag contains 4 red marbles, 3 blue marbles, and 2 green marbles. What is the probability that Nick reached into the bag and picked a red marble if you know that he did NOT pick a green marble? Did not say given that Don’t Over Think…..If he did not pick green, that means there were only 7 marbles left. His probability was 4/7. Example 4: ◦ Calculate the probability of rolling a total of 8 if the first roll is a 5. P(5) = 1/6 P(sum 8) = 3/6 P(A and B)= 3/36 Divide by 1/6 = 1/6 The expected value is often referred to as the “long-term” average or mean . This means that over the long term of doing an experiment over and over, you would expect this average. To find the expected value or long term average, simply multiply each value of the random variable by its probability and add the products. Suppose that the following game is played. A man rolls a die. If he rolls a 1, 3, or 5, he loses $3, if he rolls a 4 or 6, he loses $2, and if he rolls a 2, he wins $12. What gains or losses should he expect on average? (What is his expected value?) We must find the probabilities of each outcome. We can make a chart to help us see this. Possible Probability Losses/Gains P(losses/gains) -$3 3 6 -$2 2 6 $12 1 6 Now, we multiply the probability for each outcome by the amount of money either gained or lost for that outcome. 3 9 $3 $1.50 Expected value of rolling a 1, 3, or 5: ________________ 6 6 2 4 $2 $0.67 Expected value of rolling a 4 or 6: ________________ 6 6 1 12 $12 $2.00 Expected value of rolling a 2: _______________ 6 6 To find the expected value for the entire game (the answer), simply add up the expected value for each outcome. Expected Value = $1.50 $0.67 $2.00 $0.17 This means that the man playing this game is expected to lose an average of $0.17 each game he plays.