Statistics Chapter 5 Homework
5.2 A probability experiment is conducted in which the sample space of the experiment is S={4,5,6,7,8,9,10,11,12,13,14,15}. Let event E={4,5,6,7,8}. Assume each outcome is equally likely. List the outcomes in E^c. Find P(E^c). 1. List the outcomes in E^c. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. E^c= ___ (Use a comma to separate answers as needed.) B. E^c={ } 2. P(E^c)=_____ (Type an integer or a decimal rounded to three decimal places as needed.)
1. A. {9, 10, 11, 12, 13, 14, 15} 2. 0.583
5.5 Find the value of the combination. 5C2 5C2=_____
10
5.5 A woman has ten skirts and two blouses. Assuming that they all match, how many different skirt-and-blouse combinations can she wear? The woman can wear ____ different skirt-and-blouse combinations. (Type a whole number.)
10 x 2 = 20
5.3 The word or in probability implies that we use the ______ Rule.
Addition Note: The word or in probability implies that we use the Addition Rule.
5.2 Find the probability P(E^c) if P(E)=0.54.
Let S denote the sample space of a probability experiment and let E denote an event. The complement of E, denoted E^c, is all outcomes in the sample space S that are not outcomes in the event E. If E represents any event and E^c represents the complement of E, then the probability P(E^c) is given by the formula below. P(E^c)=1−P(E) The probability P(E^c) is 1 − 0.54 = 0.46
5.2 If E and F are disjoint events, then P(E or F)=______
P(E)+P(F) Note: If E and F are disjoint events, then P(E or F)=P(E)+P(F).
5.2 Find the probability of the indicated event if P(E)=0.4 and P(F)=0.45. Find P(E or F) if P(E and F)=0.2.
Use the general addition rule to find the probability. The rule states that for any two events E and F, P(E or F)=P(E)+P(F)−P(E and F). Substitute the values from the problem statement into the general addition rule. P(E or F) = P(E)+P(F)−P(E and F) = 0.4+0.45−0.2 Simplify the above expression to find P(E or F). P(E or F) = 0.4+0.45−0.2 =0.65
5.1 A(n) _____ is any collection of outcomes from a probability experiment.
event
5.5 A(n) ________ is an ordered arrangement of r objects chosen from n distinct objects without repetition.
permutation
5.2 Find the probability of the indicated event if P(E)=0.40 and P(F)=0.50. Find P(E or F) if P(E and F)=0.10. P(E or F)=______
P(E or F) = 0.8
5.4 Suppose that E and F are two events and that P(E and F)=0.1 and P(E)=0.4. What is P(F|E)? P(F|E)=____ (Type an integer or a decimal.)
0.25
5.2 Find the probability P(E or F) if E and F are mutually exclusive, P(E)=0.41, and P(F)=0.47. The probability P(E or F) is _____. (Simplify your answer.)
0.88
5.4 Suppose that E and F are two events and that P(E)=0.2 and P(F|E)=0.3. What is P(E and F)? P(E and F)= _______ (Simplify your answer.)
0.06
5.3 Suppose that events E and F are independent, P(E)=0.7, and P(F)=0.9. What is the P(E and F)? The probability P(E and F) is ______. (Type an integer or a decimal.)
0.63
5.5 Find the value of the factorial. 11! 11!=_____ (Type a whole number.)
39916800
5.2 Find the probability of the indicated event if P(E)=0.25 and P(F)=0.50. Find P(E and F) if P(E or F)=0.60 P(E and F)=_____ (Simplify your answer.)
0.15
5.1 Let the sample space be S={1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Suppose the outcomes are equally likely. Compute the probability of the event E={3, 10}. P(E)=____ (Type an integer or a decimal. Do not round.)
0.2
5.1 Let the sample space be S={1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Suppose the outcomes are equally likely. Compute the probability of the event E="an odd number less than 10." P(E)=_____ (Type an integer or a decimal. Do not round.)
0.5
5.1 Let the sample space be S={1, 2, 3, 4}. Suppose the outcomes are equally likely. Compute the probability of the event E="an odd number." P(E)=_____
0.5 Note: If an experiment has n equally likely outcomes and if the number of ways that an event E can occur is m, then the probability of E, P(E), is calculated using the equation below. P(E) = number of ways that E can occur / number of possible outcomes = m / n
5.4 Suppose that E and F are two events and that N(E and F)=280 and N(E)=500. What is P(F|E)? P(F|E)≈______ (Round to three decimal places as needed.)
0.56
5.2 A golf ball is selected at random from a golf bag. If the golf bag contains 7 type A balls, 9 type B balls, and 3 type C balls, find the probability that the golf ball is not a type A ball. The probability that the golf ball is not a type A ball is _____. (Type an integer or decimal rounded to three decimal places as needed.)
0.632 Note: Let S denote the sample space of a probability experiment and let E denote an event. The complement of E, denoted E^c, is all outcomes in the sample space S that are not outcomes in the event E. Note that the event that the golf ball is not a type A ball is the complement of the event that the golf ball is a type A ball. If E represents any event and E^c represents the complement of E, then the probability P(E^c) is given by the formula below. P(E^c)=1−P(E)
5.2 Find the probability P(E^c) if P(E)=0.36. The probability P(E^c) is ______
0.64
5.2 A golf ball is selected at random from a golf bag. If the golf bag contains 4 orange balls, 8 green balls, and 4 brown balls, find the probability of the following event. The golf ball is orange or green. The probability that the golf ball is orange or green is _____. (Type an integer or a decimal rounded to three decimal places as needed.)
0.75
5.4 According to a research agency, in 25% of marriages the woman has a bachelor's degree and the marriage lasts at least 20 years. According to a census report, 28% of women have a bachelor's degree. What is the probability a randomly selected marriage will last at least 20 years if the woman has a bachelor's degree? Note: 51% of all marriages last at least 20 years. The probability that a randomly selected marriage will last at least 20 years if the woman has a bachelor's degree is ______. (Round to three decimal places as needed.)
0.893
5.1 Complete the sentence below. In a probability model, the sum of the probabilities of all outcomes must equal _____. (Type an integer or a decimal. Do not round.)
1
5.5 Find the value of the permutation. 16P0 16P0=_______ (Simplify your answer.)
1
5.1 Is the following a probability model? What do we call the outcome "red"? Color: Probability red: 0 green: 0.2 blue: 0.25 brown: 0.1 yellow: 0.15 orange: 0.25 1. Is the table above an example of a probability model? A. No, because the probabilities do not sum to 1 . B. Yes, because the probabilities sum to 1 and they are all greater than or equal to 0 and less than or equal to 1. C. No, because not all the probabilities are greater than 0. D. Yes, because the probabilities sum to 1. 2. What do we call the outcome "red"? A. Impossible event B. Certain event C. Not so unusual event D. Unusual event
1) A. No, because the probabilities do not sum to 1 . Note: To determine if the table shows a probability model, recall the rules of probabilities, which are shown below. 1. The probability of any event E, P(E), must be greater than or equal to 0 and less than or equal to 1. If we let E denote any event, then 0≤P(E)≤1. 2. The sum of probabilities of all outcomes must equal 1. That is, if the sample space 2)A. Impossible event
5.4 Suppose that two cards are randomly selected from a standard 52-card deck. (a) What is the probability that the first card is a king and the second card is a king if the sampling is done without replacement? (b) What is the probability that the first card is a king and the second card is a king if the sampling is done with replacement? (a) If the sampling is done without replacement, the probability that the first card is a king and the second card is a king is _____. (Round to three decimal places as needed.) (b) If the sampling is done with replacement, the probability that the first card is a king and the second card is a king is _______. (Round to three decimal places as needed.)
a) 0.005 b) 0.006
5.3 What is the probability of obtaining ten heads in a row when flipping a coin? Interpret this probability. a) The probability of obtaining ten heads in a row when flipping a coin is______ (Round to five decimal places as needed.) b) Interpret this probability. Consider the event of a coin being flipped ten times. If that event is repeated ten thousand different times, it is expected that the event would result in ten heads about ____ time(s). (Round to the nearest whole number as needed.)
a) 0.00098. b) 10
5.3 What is the probability of obtaining nine tails in a row when flipping a coin? Interpret this probability. a) The probability of obtaining nine tails in a row when flipping a coin is _____. (Round to five decimal places as needed.) b) Interpret this probability. Consider the event of a coin being flipped nine times. If that event is repeated ten thousand different times, it is expected that the event would result in nine tails about _____ time(s). (Round to the nearest whole number as needed.)
a) 0.00195 b) 20
5.1 What is the probability of an event that is impossible? Suppose that a probability is approximated to be zero based on empirical results. Does this mean that the event is impossible? (a)What is the probability of an event that is impossible? (b) Suppose that a probability is approximated to be zero based on empirical results. Does this mean that the event is impossible? Yes No
(a) 0 (zero) (b) No Note: The correct answer is No because when a probability is based on an empirical experiment, a probability of zero does not mean that the event cannot occur. The probability of an event E is approximately the number of times event E is observed divided by the number of repetitions of the experiment, as shown below. Just because the event is not observed, does not mean that the event is impossible. P(E)≈relative frequency of E = frequency of E / number of trials of experiment
5.2 A standard deck of cards contains 52 cards. One card is selected from the deck. (a)Compute the probability of randomly selecting an eight or ten. (b)Compute the probability of randomly selecting an eight or ten or nine. (c)Compute the probability of randomly selecting a nine or diamond. (a) P(eight or ten)=_____ (Type an integer or a decimal rounded to three decimal places as needed.)
(a) 0.154 Note: Use the addition rule for disjoint events to find the probability. The rule states that if E and F are disjoint (or mutually exclusive) events, then P(E or F)=P(E)+P(F). (b) 0.231 (c) 0.308 Note: Use the general addition rule to find the probability. The rule states that for any two events E and F, P(E or F)=P(E)+P(F)−P(E and F). The probability of drawing a jack from a deck of cards is approximately 0.077, and the probability of drawing a heart is 0.25. Find the probability of drawing a jack that is also a heart from a deck of cards, rounding to three decimal places.
5.1 The table shows the results of rolling a fair six-sided die. Complete parts (a) through (d) below. Outcome on Die : 100 Trials: 100 Trials :500 Trials 1: 15: 18: 72 2: 18: 15: 93 3: 20: 19: 87 4: 15: 14: 71 5: 16: 13: 90 6: 16: 21: 87 (a) Using the table, find the empirical probability of rolling a 3 for the first 100 trials. The empirical probability of rolling a 3 for the first 100 trials is ______. (Round to two decimal places as needed.) (b) Using the table, find the empirical probability of rolling a 3 for the second 100 trials. The empirical probability of rolling a 3 for second 100 trials is ____ (Round to two decimal places as needed.) (c) Using the table, find the empirical probability of rolling a 3 for 500 trials. The empirical probability of rolling a 3 for 500 trials is ____ (Round to two decimal places as needed.) (d) Compare the empirical probabilities to the probability obtained using the classical method, and explain what they show. Choose the correct answer below. A. The empirical probabilities are all equally far away from the classical probability. This is an outcome expected according to the Law of Large Numbers. B. All the empirical probabilities are equal to the classical probability. This is an outcome expected according to the Law of Large Numbers. C. The empirical probability is closest to the classical probability for the two sets of 100 trials and gets farther away for the 500 trials. This is an outcome expected according to the Law of Large Numbers. D. The empirical probabilities approach the classical probability as the sample size increases. This is an outcome expected according to the Law of Large Numbers.
(a) 0.20 (b)0.19 (c)0.17 (d)D. The empirical probabilities approach the classical probability as the sample size increases. This is an outcome expected according to the Law of Large Numbers Note:First, calculate the probabilities using the classical method and then compare the empirical probabilities for each trial size to the calculated classical probability. Remember that the classical probability is the number of outcomes in E divided by the number of outcomes in the sample space. P(E)=Number of ways that E can occur / Number of possible outcomes=m/n
5.1 A survey of 800 randomly selected high school students determined that 185 play organized sports. (a) What is the probability that a randomly selected high school student plays organized sports? (b) Interpret this probability. (a) The probability that a randomly selected high school student plays organized sports is ____ (Round to the nearest thousandth as needed.) (b) Choose the correct answer below. (Type a whole number.) A. If 1,000 high school students were sampled, it would be expected that about ____ of them play organized sports. B. If 1,000 high school students were sampled, it would be expected that exactly _____ of them play organized sports.
(a) 0.231 (b) 231
5.1 A baseball player hit 67 home runs in a season. Of the 67 home runs, 18 went to right field, 23 went to right center field, 12 went to center field, 12 went to left center field, and 2 went to left field. (a) What is the probability that a randomly selected home run was hit to right field? (b) What is the probability that a randomly selected home run was hit to left field? (c) Was it unusual for this player to hit a home run to left field? Explain. (a) The probability that a randomly selected home run was hit to right field is ____ (Round to three decimal places as needed.) (b) The probability that a randomly selected home run was hit to left field is ____ (Round to three decimal places as needed.) (c) Was it unusual for this player to hit a home run to left field? A. Yes, because P(left field)<0.05. B. No, because the probability of an unusual event is 0. C. No, because this player hit 2 home runs to left field. D. Yes, because P(left field)<0.5.
(a) 0.269 (b) 0.03 (c) A. Yes, because P(left field)<0.05.
5.2 A standard deck of cards contains 52 cards. One card is selected from the deck. (a)Compute the probability of randomly selecting a diamond or club. (b)Compute the probability of randomly selecting a diamond or club or heart. (c)Compute the probability of randomly selecting a nine or diamond. (a) P(diamond or club)=____ (Type an integer or a decimal rounded to three decimal places as needed.) (b) P(diamond or club or heart)=_____ (Type an integer or a decimal rounded to three decimal places as needed.) (c) P(nine or diamond)=_____ (Type an integer or a decimal rounded to three decimal places as needed.)
(a) 0.5 (b) 0.75 (c) 0.308
5.1 A survey of 700 randomly selected high school students determined that 468 play organized sports. (a) What is the probability that a randomly selected high school student plays organized sports? (b) Interpret this probability.
(a) To compute the probability of the event E use the empirical method because a probability experiment is run. The probability of an event E is approximately the number of times event E is observed divided by the number of repetitions of the experiment. P(E)=relative frequency of E = frequency of E / number of trials of experiment Let event E="a randomly selected high school student plays organized sports." To find P(E), first find the frequency of E. The frequency of E is the number of times E has occurred. Since there are 468 high school students that play organized sports, the frequency of E is 468. Now find the number of trials of the experiment. In this case, the number of trials of the experiment is the number of students surveyed. Since the number of high school students surveyed is 700, the number of trials of the experiment is 700. Substitute 468 for the frequency of E and 700 for the number of trials of the experiment in the formula for the relative frequency of E. P(E) = frequency of E / number of trials of experiment = 468 / 700 Compute P(E), rounding to the nearest hundredth. P(E) = 468 / 700 = 0.669 Therefore, the probability that a randomly high school student plays organized sports is 0.669. (b) Probability deals with experiments that yield random short-term results or outcomes yet reveal long-term predictability. The long term proportion in which a certain outcome is observed is the probability of that outcome. Thus, if 1000 high school students were sampled, it would be expected that about 669 of them play organized sports.
5.1 example A baseball player hit 65 home runs in a season. Of the 65 home runs, 19 went to right field, 23 went to right center field, 9 went to center field, 11 went to left center field, and 3 went to left field. (a) What is the probability that a randomly selected home run was hit to right field? (b) What is the probability that a randomly selected home run was hit to left field? (c) Was it unusual for this player to hit a home run to left field? Explain.
(a) To compute the probability of the event E, use the classical method because the outcomes are equally likely. (b) If an experiment has n equally likely outcomes and if the number of ways that an event E can occur is m, then the probability of E, P(E) is calculated using the equation below. P(E) = Number of ways that E can occur / Number of possible outcomes = m / n So, if S is the sample space of the experiment, then P(E) can be calculated using the equation below, where m is the number of outcomes in E, and n is the number of outcomes in the sample space. P(E) = m / n Let event E="randomly selected home run was hit to right field." To find P(E), first find m. Recall that m is number of ways that event E can occur. Since this player hit 19 home runs to right field, m=19. Now find n. Remember that n is the number of equally likely outcomes of an experiment. When randomly selecting a home run, each of the 65 home runs is equally likely to be chosen, so n=65. Substitute 19 for the number of outcomes in E and 65 for the number of outcomes in the sample space in the formula for the probability of E. P(E) = m / n = 19 / 65 Compute P(E), rounding to three decimal places. P(E) = 19 / 65 = 0.292 Therefore, the probability that a randomly selected home run was hit to right field is 0.292. Let event E="randomly selected home run was hit to left field." To find P(E), first find m. Recall that m is the number of ways that event E can occur. Since this player hit 3 home runs to left field, m=3. Since S is the same as in part a), n=65. Substitute 3 for the number of outcomes in E and 65 for the number of the outcomes in the sample space in the formula for the probability of E. P(E) = m / n = 3 / 65 Compute P(E), rounding to three decimal places. P(E) = 3 / 65 = 0.046 Therefore, the probability that a randomly selected home run was hit to left field is 0.046. (c) An event is unusual if it has a low probability of occurring. The choice of a cutoff should consider the context of the problem. To determine if an outcome is unusual, recall that an unusual event is an event that has a low probability of occurring. Typically, an event with a probability less than 5% is considered unusual. Note that the probability of this player hitting a home run to left field is 0.046<0.05. Thus, P(left field) is an unusual event.
5.2 A standard deck of cards contains 52 cards. One card is selected from the deck. (a) Compute the probability of randomly selecting a spade or club. (b) Compute the probability of randomly selecting a spade or club or heart. (c) Compute the probability of randomly selecting a jack or heart.
(a) Use the addition rule for disjoint events to find the probability. The rule states that if E and F are disjoint (or mutually exclusive) events, then P(E or F)=P(E)+P(F). This gives P(spade or club)=P(spade)+P(club). Recall that in a deck of cards there are four of each type of card (two, three, four, five, six, seven, eight, nine, ten, jack, queen, king, and ace). These cards are divided into four suits, with one in each suit (hearts, diamonds, clubs, and spades). To find the probability of drawing a spade, first find N(spade), the number of spades, and N(S), the sample size. N(spade) = 13 N(S) = 52 Find the probability of drawing a spade. P(spade) = N(spade) / N(S) = 13 / 52 = 0.25 Now find the probability of drawing a club, where N(club) is the number of clubs. P(club) = N(club) / N(S) = 13 / 52 = 0.25 Add the two probabilities above to find the probability of randomly selecting a spade or club. P(spade or club) = 0.25+0.25 = 0.5 (b) The addition rule for disjoint events can be extended to more than two disjoint events. P(E or F or G...)=P(E)+P(F)+P(G)+... P(spade or club or heart)=P(spade)+P(club)+P(heart) The probabilities P(spade) and P(club) were calculated in part (a). Find the probability of drawing a heart, where N(heart) is the number of hearts. P(heart) = N(heart) / N(S) = 13 / 52 = 0.25 Add the three probabilities above to find the probability of randomly selecting a spade or club or heart. P(spade or club or heart) = 0.25+0.25+0.25 = 0.75 (c) Use the general addition rule to find the probability. The rule states that for any two events E and F, P(E or F)=P(E)+P(F)−P(E and F). P(jack or heart)=P(jack)+P(heart)−P(jack and heart) The probability of drawing a jack from a deck of cards is approximately 0.077, and the probability of drawing a heart is 0.25. Find the probability of drawing a jack that is also a heart from a deck of cards, rounding to three decimal places. P(jack and heart) = N(jack and heart) / N(S) = 1 / 52 = 0.019 Use the three probabilities above to find the probability of randomly selecting a jack or heart. P(jack or heart) = 0.077+0.25−0.019 = 0.308
5.1 In a national survey college students were asked, "How often do you wear a seat belt when riding in a car driven by someone else?" The response frequencies appear in the table to the right. (a) Construct a probability model for seat-belt use by a passenger. (b) Would you consider it unusual to find a college student who never wears a seat belt when riding in a car driven by someone else? Response: Frequency Never: 127 Rarely: 337 Sometimes: 574 Most of the time: 1098 Always: 2541
(a) When a probability experiment is run, probabilities are approximated using the empirical approach. That is, the probability of an event E is approximately the number of times event E is observed divided by the number of repetitions of the experiment. P(E)≈relative frequency of E = frequency of E / number of trials of experiment To construct a probability model, use the equation above. First determine the number of trials of this experiment. Note that the individual students can be thought of as trials of the experiment. number of trials = 127+337+574+1098+2541 = 4677 Now for each response, divide the frequency by the number of trials of the experiment to find the approximate probability. Response: Probability Never: 127/4677=0.027 Rarely: 337/4677=0.072 Sometimes: 574/4677=0.123 Most of the time: 1098/4677=0.235 Always: 2541/4677=0.543 Thus, the probability model for seat-belt use by a passenger is shown below. Response: Frequency: Probability Never: 127: 0.027 Rarely: 337: 0.072 Sometimes: 574: 0.123 Most of the time: 1098: 0.235 Always: 2541: 0.543 (b) To determine if a response of "never" is unusual, recall that an unusual event is an event that has a low probability of occurring. Typically, an event with a probability less than 5% is considered unusual. Recall from part a that the probability of finding a college student who never wears a seat belt when riding in a car driven by someone else is 0.027. Use this value to determine if this event is unusual.
5.1 In a national survey college students were asked, "How often do you wear a seat belt when riding in a car driven by someone else?" The response frequencies appear in the table to the right. (a) Construct a probability model for seat-belt use by a passenger. (b) Would you consider it unusual to find a college student who never wears a seat belt when riding in a car driven by someone else? Response: Frequency Never 128 Rarely 338 Sometimes 587 Most of the time 1289 Always 2693 (a) Complete the table below. Response:Probability Never: ____ (Round to the nearest thousandth as needed.) Rarely: _____ (Round to the nearest thousandth as needed.) Sometimes: ______ (Round to the nearest thousandth as needed.) Most of the time: ____ (Round to the nearest thousandth as needed.) Always: ____ (Round to the nearest thousandth as needed.) (b) Would you consider it unusual to find a college student who never wears a seat belt when riding in a car driven by someone else? A. No, because there were 128 people in the survey who said they never wear their seat belt. B. Yes, because 0.01<P(never)<0.10. C. Yes, because P(never)<0.05. D. No, because the probability of an unusual event is 0.
(a)Response: Probability Never 0.025 (Round to the nearest thousandth as needed.) Rarely 0.067 (Round to the nearest thousandth as needed.) Sometimes 0.117 (Round to the nearest thousandth as needed.) Most of the time 0.256 (Round to the nearest thousandth as needed.) Always 0.535 (Round to the nearest thousandth as needed.) (b)C. Yes, because P(never)<0.05. Note: To determine if a response of "never" is unusual, recall that an unusual event is an event that has a low probability of occurring. Typically, an event with a probability less than 5% is considered unusual.
5.2 A probability experiment is conducted in which the sample space of the experiment is S={2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}, event E={3, 4, 5, 6, 7} and event G={9, 10, 11, 12}. Assume that each outcome is equally likely. List the outcomes in E and G. Are E and G mutually exclusive? 1. List the outcomes in E and G. Choose the correct answer below. A. E and G={___} (Use a comma to separate answers as needed.) B. E and G={ } 2. Are E and G mutually exclusive? A. No, because the events E and G have at least one outcome in common. B. Yes, because the events E and G have no outcomes in common. C. Yes, because the events E and G have at least one outcome in common. D. No, because the events E and G have outcomes in common.
1) B. E and G={ } 2) B. Yes, because the events E and G have no outcomes in common.
5.2 A probability experiment is conducted in which the sample space of the experiment is S={12,13,14,15,16,17,18,19,20,21,22,23}. Let event E={13,14,15,16,17,18} and event F={17,18,19,20}. List the outcomes in E and F. Are E and F mutually exclusive? 1. List the outcomes in E and F. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. ____ (Use a comma to separate answers as needed.) B.{___} 2. Are E and F mutually exclusive? A. Yes. E and F have no outcomes in common. B. Yes. E and F have outcomes in common. C. No. E and F have outcomes in common. D. No. E and F have no outcomes in common.
1. a. 17,18 2. C. No. E and F have outcomes in common. Note: Two events are disjoint if they have no outcomes in common. Another name for disjoint events is mutually exclusive events.
5.5 Suppose Jim is going to build a playlist that contains 14 songs. In how many ways can Jim arrange the 14 songs on the playlist? Jim can arrange the 14 songs on the playlist in _____ different ways.
14P14 = 87178291200 Note: Determine if the choices are distinct, if the order of the choices is important, and if the choices can be repeated. Use this information to choose the correct method of counting.
5.5 A lottery exists where balls numbered 1 to 18 are placed in an urn. To win, you must match the five balls chosen in the correct order. How many possible outcomes are there for this game? The number of possible outcomes is ______. (Simplify your answer.)
18P5 = 1028160
5.5 Find the value of the permutation. 2P1 2P1=______ (Type a whole number.)
2
5.5 Find the value of the factorial. 4! = _____
24 Note: On TI-30XS Multiview Calculator Insert 4 [prb] then go to 3 for ! and enter
5.5 Suppose 31 cars start at a car race. In how many ways can the top 3 cars finish the race? The number of different top three finishes possible for this race of 31 cars is _____. (Use integers for any number in the expression.)
31P3= 26970
5.5 How many different simple random samples of size 5 can be obtained from a population whose size is 38? The number of simple random samples which can be obtained is ______. (Type a whole number.)
38C5 = 51942
5.5 Find the value of the combination. 4C3 4C3=_____
4
5.5 Find the value of the permutation. 8P7 8P7=_____ (Type a whole number.)
40320
5.5 Find the value of the permutation. 8P8 8P8=_____ (Type a whole number.)
40320
5.1 Why is the following not a probability model? Click the icon to view the data table. Determine why it is not a probability model. Choose the correct answer below. A. This is not a probability model because at least one probability is less than 0. B. This is not a probability model because the sum of the probabilities is not 1. C. This is not a probability model because at least one probability is greater than 1. D. This is not a probability model because at least one probability is greater than 0.
A. This is not a probability model because at least one probability is less than 0.
5.3 What is the probability of obtaining eight tails in a row when flipping a coin? Interpret this probability.
Assume that the coin is a fair coin. Two events E and F are independent if the occurrence of event E in a probability experiment does not affect the probability of event F. Two events are dependent if the occurrence of event E in a probability experiment affects the probability of event F. Since the probability of obtaining a tail on one toss, does not have any effect on the likelihood of obtaining a tail on another toss, these events are independent. Determine the probability of obtaining tails on a single flip. P(tail)=1/2 Use the multiplication rule for n independent events to find the probability. If events E, F, G, ... are independent events, then the probability P(E and F and G and ...) is given by the formula below. P(E and F and G and ...)=P(E)•P(F)•P(G)•... Find the probability of obtaining eight tails in a row when flipping a coin, rounding to five decimal places. P(tails on all eight flips) = P(tail on 1st flip)•...•P(tail on 8th flip) = 1/2•1/2•1/2•1/2•1/2•1/2•1/2•1/2 = 0.00391 The probability of an event E is approximately the number of times event E is observed divided by the number of repetitions of the experiment. P(E)≈relative frequency of E=frequency of E / number of trials of experiment Assume that E is the event of obtaining eight tails in a row when flipping a coin. If the number of trials of the experiment is 10,000, the expected frequency of E is 10,000•P(E). Interpret the probability P(E)=0.00391. Consider the event of a coin being flipped eight times. If that event is repeated ten thousand different times, it is expected that the event would result in eight tails about 10,000•0.00391≈39 times.
5.5 A license plate is to consist of 3 digits followed by 2 uppercase letters. Determine the number of different license plates possible if the first and second digits must be odd, and repetition is not permitted. Choose the correcrt answer below. A. 58,240,000 B. 104,000 C. 810 D. 5,241,600,000
B. 104,000
5.5 Suppose that a local area network requires five letters for user names. Lower- and uppercase letters are considered the same. How many user names are possible for the local area network? How many user names are possible for the local area network? A. 5! B. 26^5 C. 26 D. 5
B. 26^5
5.5 Suppose Bill is going to build a playlist that contains 15 songs. In how many ways can Bill arrange the 15 songs on the playlist?
Before proceeding you must assess the situation to decide what you are counting. Since the songs are distinct, no song can be repeated on the playlist, and order is important, you should count permutations. The number of arrangements of r objects chosen from n objects, in which 1. the n objects are distinct, 2. once an object is used, it cannot be repeated, and 3. order is important, is given by the following formula. nPr = n! / (n−r)! Since order is important on the playlist and Bill would like to use all 15 songs, the number of arrangements which can be made is as follows. 15P15 = 15! / (15−15)! = 15! = 1,307,674,368,000 Bill can arrange the 15 songs on the playlist in 1,307,674,368,000 different ways.
5.1 What does it mean for an event to be unusual? Why should the cutoff for identifying unusual events not always be 0.05? Choose the correct answer below. A. An event is unusual if it has a probability equal to 0. The choice of a cutoff should consider the context of the problem. B. An event is unusual if it has a probability equal to 1. The choice of a cutoff should consider the context of the problem. C. An event is unusual if it has a low probability of occurring. The choice of a cutoff should consider the context of the problem. D. An event is unusual if it has a high probability of occurring. The choice of the cutoff should consider the context of the problem.
C. An event is unusual if it has a low probability of occurring. The choice of a cutoff should consider the context of the problem.
5.2 According to a center for disease control, the probability that a randomly selected person has hearing problems is 0.156. The probability that a randomly selected person has vision problems is 0.082. Can we compute the probability of randomly selecting a person who has hearing problems or vision problems by adding these probabilities? Why or why not? Choose the correct answer below. A. No, because hearing problems and vision problems are events that are too similar to one another. B. Yes, because hearing and vision are two different senses, and therefore, they are two unique problems. C. No, because hearing and vision problems are not mutually exclusive. So, some people have both hearing and vision problems. These people would be included twice in the probability. D. Yes, because this is an application of the Addition Rule for Disjoint Events.
C. No, because hearing and vision problems are not mutually exclusive. So, some people have both hearing and vision problems. These people would be included twice in the probability. Note: Some people have both hearing and vision problems. This means that the events of having a hearing problem and having a vision problem are not disjoint (or not mutually exclusive). Remember, when two events are not disjoint the General Addition Rule must be used. The General Addition Rule states that for any two events E and F, P(E or F)=P(E)+P(F)−P(E and F). That is, the probability is found by adding together the two given probabilities and subtracting the probability of a person having both problems. If the probability of a person having both problems is not subtracted out, then these people will be included twice in the probability.
5.5 A license plate is to consist of 5 digits followed by 2 uppercase letters. Determine the number of different license plates possible if the first and second digits must be odd, and repetition is not permitted. Choose the correcrt answer below. A. 74,880,000 B. 6,739,200,000 C. 7,370 D. 4,368,000
D. 4,368,000 Note: The Multiplication Rule of Counting states that if a task consists of a sequence of choices in which there are p selections for the first choice, q selections for the second choice, r selections for the third choice, and so on, then the task of making these selections can be done in p•q•r•... different ways. Check that you have correctly determined the possible choices for each digit and letter.
5.4 The notation P(F E) means the probability of event _____ given event ____
F; E Note: The notation P(F E) means the probability of event F given event E.
5.5 A man has five shirts and two ties. Assuming that they all match, how many different shirt-and-tie combinations can he wear?
If a task consists of a sequence of choices in which there are p selections for the first choice and q selections for the second choice, then the task of making these selections can be done in p•q different ways. In this case, the man can choose a shirt 5 different ways. After he chooses a shirt, the man can choose a tie 2 different ways. Now we calculate the number of choices the man has for shirt-and-tie combinations. 5•2=10 The man can wear 10 different shirt-and-tie combinations.
5.5 Suppose that a local area network requires three letters for user names. Lower- and uppercase letters are considered the same. How many user names are possible for the local area network?
If a task consists of a sequence of choices in which there are p selections for the first choice, q selections for the second choice, r selections for the third choice, and so on, then the task of making these selections can be done in p•q•r•... different ways. Recall that there are 26 letters in the alphabet and that lower- and uppercase letters are considered the same. Determine how many different choices there are for the first letter. p=26 Determine how many different choices there are for the second letter. q=26 Thus, there are 26 ways to choose the first letter, 26 ways to choose the second letter, 26 ways to choose the third letter. By the multiplication rule, the number of user names that are possible for the local area network is 26•26•26 or 26^3.
5.5 Find the value of the factorial. 9!
If n≥0 is an integer, the factorial symbol n! is defined below. 0!=1 1!=1 n!=n•(n−1)•...•3•2•1 To compute the factorial it may help to list the terms of the product. Here, we have the following product to consider. 9! = 9•8•7•6•5•4•3•2•1 Now multiply. 9! = 9•8•7•6•5•4•3•2•1 = 362,880
5.2 A golf ball is selected at random from a golf bag. If the golf bag contains 4 type A balls, 10 type B balls, and 5 type C balls, find the probability that the golf ball is not a type A ball.
Let S denote the sample space of a probability experiment and let E denote an event. The complement of E, denoted Ec, is all outcomes in the sample space S that are not outcomes in the event E. Note that the event that the golf ball is not a type A ball is the complement of the event that the golf ball is a type A ball. Let A be the event that the golf ball is a type A ball. Then Ac is the event that the selected golf ball is not a type A ball. If S is the sample space of this experiment, the probability of A is given by the formula, where N(A) is the number of outcomes in A, and N(S) is the number of outcomes in the sample space. P(A)=N(A) / N(S) Find the number of outcomes in A. Notice that the event A is selecting a type A golf ball. N(A)=4 Find N(S), the number of outcomes in the sample space. N(S) = 4+10+5 = 19 Now find the probability P(A), rounded to three decimal places. P(A) = N(A) / N(S) = 4 / 19 = 0.211 If E represents any event and E^c represents the complement of E, then the probability P(E^c) is given by the formula below. P(E^c)=1−P(E) Now find P(A^c), the probability that the selected golf ball is not a type A golf ball. P(A^c) = 1−P(A) = 1−0.211 = 0.789 The probability that the selected golf ball is not a type A ball is 0.789.
5.4 The probability that a randomly selected individual in a country earns more than $75,000 per year is 9.5%. The probability that a randomly selected individual in the country earns more than $75,000 per year, given that the individual has earned a bachelor's degree, is 21.5%. Are the events "earn more than $75,000 per year" and "earned a bachelor's degree" independent? Are these events independent? Yes or No
No Note: The answer is No because the probability that a randomly selected individual in the country earns more than $75,000 per year, given that the individual has earned a bachelor's degree, is not equal to the probability that a randomly selected individual in a country earns more than $75,000 per year. Two events E and F are independent if P(E F)=P(E) or, equivalently, if P(F E) = P(F).
5.2 Fill in the blank. If E and F are not disjoint events, then P(E or F)=________.
P(E)+P(F)−P(E and F) Note: If the addition rule for disjoint events P(E or F)=P(E)+P(F) is used to compute the probability, the outcomes that E and F have in common will be double counted, and so P(E and F) must be subtracted if E and F are not disjoint. Hence, the general addition rule is used, which states that for any two events E and F, P(E or F)=P(E)+P(F)−P(E and F).
5.5 How many different simple random samples of size 9 can be obtained from a population whose size is 49?
Since the individuals are distinct, the order does not matter, and an individual cannot be selected more than once for each sample, we are counting combinations in this situation. A combination is the selection of r objects from a set of n different objects when the order in which the objects is selected does not matter (so AB is the same as BA) and an object cannot be selected more than once (repetition is not allowed). The number of combinations is given by the following formula. nCr = n! / r!(n−r!) The value of n is the number of people to choose from, so n=49. The value of r is the number of people to choose, so r=9. Substitute the numbers into the formula for combinations and simplify. 49C9 =49! / 9!(49−9)! =49•48•47•46•45•44•43•42•41•40! / 9!•40! =49•48•47•46•45•44•43•42•41 / 9•8•7•6•5•4•3•2•1 =745,520,860,465,920 / 362,880 =2,054,455,634 The number of simple random samples which can be obtained is 2,054,455,634.
5.5 A license plate is to consist of 4 digits followed by 4 uppercase letters. Determine the number of different license plates possible if the first and second digits must be odd, and repetition is not permitted.
The Multiplication Rule of Counting states that if a task consists of a sequence of choices in which there are p selections for the first choice, q selections for the second choice, r selections for the third choice, and so on, then the task of making these selections can be done in p•q•r•... different ways. The first digit in the license plate must be odd. There are 5 odd digits from 0-9 so there are 5 possible choices for the first digit. The second digit on the license plate must also be odd, and repetition is not allowed. Since there are 5 odd digits, and one digit is used for the first choice, there are 4 possible choices for the second digit. The third digit on the license plate can be either even or odd, but repetition is still not allowed. Since repetition is not allowed and 2 digits have already been used, that leaves 8 possible choices for the third digit. The number of possible choices for the fourth digit is found in the same manner. The possible number of choices for each digit is as shown. 5/D 4/D 8/D 7/D Then consider the 4 letters in the code. Since there are 26 letters in the alphabet, there are 26 possible choices for the first letter. Repetition is not allowed, so there are 25 choices for the second letter. The possible number of choices for each digit as well as each letter is as shown. 5/D 4/D 8/D 7/D 26/L 25/L 24/L 23/L Use the Multiplication Rule of Counting to determine the number of possible license plates. 5•4•8•7•26•25•24•23=401,856,000 Thus, there are 401,856,000 possible different license plates.
5.2 A probability experiment is conducted in which the sample space of the experiment is S={11,12,13,14,15,16,17,18,19,20,21,22}. Let event E={12,13,14,15,16,17,18}. Assume each outcome is equally likely. List the outcomes in E^c. Find P(E^c).
The complement of E, denoted E^c, is all outcomes in the sample space S that are not outcomes in the event E. Notice that the outcomes in the sample space S are given in the problem statement along with the outcomes in event E. Notice that there are outcomes in the sample space S that are not in event E because there are 7 outcomes listed in event E and 12 outcomes listed in the sample space S. This means there are some outcomes in the sample space S that are not in event E. Determine the outcomes listed in the sample space S that are not listed in event E. The outcomes are {11, 19, 20, 21, 22}. Thus, the outcomes in Ec are {11, 19, 20, 21, 22}. To find PEc, divide the number of outcomes in Ec by the number of outcomes in S. First find N(E^c). N(E^c)=5 Now find N(S). N(S)=12 Now find P(E^c), rounded to three decimal places. P(E^c) = N(E^c) / N(S) = 5 / 12 = 0.417
5.4 Suppose there is a 25.6% probability that a randomly selected person aged 50 years or older is a smoker. In addition, there is a 15.6% probability that a randomly selected person aged 50 years or older is male, given that he or she smokes. What is the probability that a randomly selected person aged 50 years or older is male and smokes? Would it be unusual to randomly select a person aged 50 years or older who is male and smokes?
The general multiplication rule states that the probability that two events E and F both occur is the probability of event E occurring times the probability of event F occurring, given the occurrence of E. P(E and F)=P(E)•P(F|E) Therefore, the probability that a randomly selected person 50 years or older is male and smokes can be found using the rule below. P(male who smokes)=P(person who smokes)•P(male | person who smokes) There is a 25.6% probability that a randomly selected person aged 50 years or older is a smoker. Convert this percent to a decimal. P(person who smokes)=0.256 There is a 15.6% probability that a randomly selected person aged 50 years or older is male, given that he or she smokes. Convert this percent to a decimal. P(male | person who smokes) = 0.156 Multiply to find the probability. P(male who smokes) = P(person who smokes)•P(male | person who smokes) = 0.256•0.156 ≈ 0.040 An event is considered unusual if its probability is less than 5%. The probability is less than 5%, so it would be considered unusual.
5.2 example A probability experiment is conducted in which the sample space of the experiment is S={13,14,15,16,17,18,19,20,21,22,23,24}. Let event E={14,15,16,17,18,19} and event F={21,22,23,24}. List the outcomes in E and F. Are E and F mutually exclusive?
The list of outcomes in E and F are the outcomes that event E and event F have in common. Notice that the outcomes in event E are {14,15,16,17,18,19} and the outcomes in event F are {21,22,23,24}. These two events have no outcomes in common. Since there are no outcomes in common in E and F, the list of outcomes in E and F is { }. Two events are disjoint if they have no outcomes in common. Another name for disjoint events is mutually exclusive events. Since E and F have no outcomes in common, E and F are mutually exclusive.
5.4 Suppose that E and F are two events and that P(E and F)=0.15 and P(E)=0.8. What is P(F|E)?
The notation P(F|E) is read "the probability of event F given event E." It is the probability that an event F occurs, given that the event E has occurred. If E and F are any two events, then the probability of event F occurring, given the occurrence of event E, is found by dividing the probability of E and F by the probability of E. All probability values are numbers between 0 and 1. P(F|E)=P(E and F) / P(E)=N(E and F) / N(E) Apply the definition for P(F|E). P(F|E)=P(E and F) / P(E)=0.15 / 0.8 Then divide to find the probability. 0.15 / 0.8 = 0.1875 Therefore, the probability is P(F|E) = 0.1875.
5.5 Find the value of the permutation. 13P7
The number of arrangements of r objects chosen from n objects, in which 1. the n objects are distinct, 2. once an object is used, it cannot be repeated, and 3. order is important, is given by the following formula. nPr = n! / (n−r)! Now, use the formula to find the value of the permutation. Begin by substituting the correct numbers in the permutation formula. 13P7 = 13! / (13−7)! Evaluate the subtraction in the denominator. 13P7 = 13! / 6! Expand the factorial symbols and cancel the factors that the denominator and numerator share. 13P7 = (13•12•11•10•9•8•7•6!) / 6! Evaluate the remaining product. This is the value of the permutation. 13P7 = 8,648,640
5.4 Suppose that E and F are two events and that P(E)=0.8 and P(F|E)=0.7. What is P(E and F)?
The probabilities P(E), P(F|E), and P(E and F) are related by the conditional probability rule shown below. P(F|E)=P(E and F) / P(E) Substitute the known values into the conditional probability rule. 0.7=P(E and F) / 0.8 Then solve the equation for P(E and F). P(E and F)=0.56 Therefore, when P(E)=0.8 and P(F|E)=0.7, the probability P(E and F) is 0.56.
5.4 According to a research agency, in 20% of marriages the woman has a bachelor's degree and the marriage lasts at least 20 years. According to a census report, 46% of women have a bachelor's degree. What is the probability a randomly selected marriage will last at least 20 years if the woman has a bachelor's degree? Note: 53% of all marriages last at least 20 years.
The probability that an event F occurs, given that the event E has occurred, is called a conditional probability and is written P(F|E). E is the event which is known to have occurred. The question asks for the probability that the selected marriage lasts at least 20 years if the woman has a bachelor's degree, so it is known that the woman in the selected marriage has a bachelor's degee. F is the event for which the probability is sought. The question asks for the probability that the selected marriage lasts at least 20 years. If E and F are any two events, then the probability of event F occurring, given the occurrence of event E, is found by dividing the probability of E and F by the probability of E, or dividing the number of outcomes in E and F by the number of outcomes in E. P(F|E)=P(E and F) / P(E)=N(E and F) / N(E) The probability of event E and F is the probability that the woman has a bachelor's degree and the marriage lasts at least 20 years. Therefore, the probability that a randomly selected marriage lasts at least 20 years if the woman has a bachelor's degree can be found as shown below. P(F|E)=P(E and F) / P(E)=P(woman has a bachelor's degree and marriage lasts at least 20 years) / P(woman has a bachelor's degree) Apply the definition for P(F|E). Notice that the percentages have been converted to probabilities. P(F|E)=P(woman has a bachelor's degree and marriage lasts at least 20 years) / P(woman has a bachelor's degree) = 0.20 / 0.46 Then divide to find the probability, rounding to three decimal places. 0.20 / 0.46 ≈ 0.435 Therefore, the probability that a randomly selected marriage will last at least 20 years if the woman has a bachelor's degree is approximately 0.435.
5.4 Suppose that two cards are randomly selected from a standard 52-card deck. (a) What is the probability that the first card is a heart and the second card is a heart if the sampling is done without replacement? (b) What is the probability that the first card is a heart and the second card is a heart if the sampling is done with replacement?
The probability that two events E and F both occur is the probability of event E occurring times the probability of event F occurring, given the occurrence of event E. P(E and F)=P(E)•P(F E) Here E is the event "the first card is a heart" and F is the event "the second card is a heart." (a) When sampling without replacement, once an object is selected, it is removed from the population and cannot be chosen again. If the sampling is done without replacement, the events "the first card is a heart" and "the second card is a heart" are dependent because the occurrence of event E affects the probability of event F The probability that the first card is a heart is the number of hearts in a deck divided by the total number of cards in a deck. The number of hearts in a deck is 13. The total number of cards in a deck is 52. Thus, the probability that the first card is a heart is 13 / 52, or 1 / 4. The probability that the second card is a heart, given that the first card is a heart, is the number of hearts left in a deck after the first selection of a heart divided by the number of cards left in a deck after the first selection. After the first selection of a heart, the number of hearts is 12 and the total number of cards is 51. Thus, the probability that the second card is a heart, given that the first card is a heart, is 12 / 51, or 4 / 17. Substitute these probabilities into the general multiplication rule and find the probability that the first card is a heart and the second card is a heart if the sampling is done without replacement, rounding to three decimal places. P(E and F) = P(E)•P(F E) = 1 / 4 • 4 / 17 = 0.059 (b) Sampling with replacement means that the selected object is placed back into the population and so could be chosen a second time. If the sampling is done with replacement, the events "the first card is a heart" and "the second card is a heart" are independent because the occurrence of event E does not affect the probability of event F. Thus, use the multiplication rule for independent events. P(E and F)=P(E)•P(F) The probabilities P(E) and P(F) are the same and equal to 1 / 4 as found above. Find the probability that the first card is a heart and the second card is a heart if the sampling is done with replacement, rounding to three decimal places. P(E and F) = P(E)•P(F) = 1 / 4 • 1 / 4 = 0.063 Thus, if the sampling is done without replacement, the probability that the first card is a heart and the second card is a heart is 0.059, and if the sampling is done with replacement, the probability is 0.063.
5.1 example Let the sample space be S={1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Suppose the outcomes are equally likely. Compute the probability of the event E={2, 3, 4, 8, 9}.
To compute the probability of the event E, use the classical method because the outcomes are equally likely. If an experiment has n equally likely outcomes and if the number of ways that an event E can occur is m, then the probability of E, P(E) is calculated using the equation below. P(E) = Number of ways that E can occur / Number of possible outcomes = m / n So, if S is the sample space of the experiment, then P(E) can be calculated using the equation below, where N(E) is the number of outcomes in E, and N(S) is the number of outcomes in the sample space. P(E) = N(E) / N(S) First find N(E). N(E) is the number of outcomes in E={2, 3, 4, 8, 9}. Thus, N(E)=5. Now find N(S). N(S) is the number of outcomes in the sample space S={1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Thus, N(S)=10. Substitute 5 for the number of outcomes in E and 10 for the number of outcomes in the sample space in the formula for the probability of E. P(E) = N(E) / N(S) = 5 / 10 Finally, compute P(E). P(E) = 5 / 10 = 0.5 Therefore, the probability of the event E={2, 3, 4, 8, 9} is 0.5.
5.1 Let the sample space S={1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}. Suppose the outcomes are equally likely. Compute the probability of the event E="an even number."
To compute the probability of the event E, use the classical method because the outcomes are equally likely. If an experiment has n equally likely outcomes and if the number of ways that an event E can occur is m, then the probability of E, P(E) is calculated using the equation below. P(E)=Number of ways that E can occur / Number of possible outcomes = m / n So, if S is the sample space of the experiment, then P(E) can be calculated using the equation below, where N(E) is the number of outcomes in E, and N(S) is the number of outcomes in the sample space. P(E)=N(E) / N(S) To determine N(E), first write out event E="an even number" for when the sample space S={1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}. Event E={2, 4, 6, 8, 10, 12} Now find N(E) and N(S). N(E) is the number of outcomes in E={2, 4, 6, 8, 10, 12}. Thus, N(E)=6. N(S) is the number of outcomes in the sample space S={1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}. Thus, N(S)=12. Substitute 6 for the number of outcomes in E and 12 for the number of outcomes in the sample space in the formula for the probability of E. P(E) = N(E)N(S) = 6 / 12 Compute P(E). P(E) = 6 / 12 = 0.5 Therefore, the probability of the event E="an even number" is 0.5.
5.5 Determine if the following statement is true or false. In a combination problem, order is not important. Choose the correct answer below. True or False
True Note: A combination is the selection of r objects from a set of n different objects when the order in which the objects are selected does not matter (so AB is the same as BA) and an object cannot be selected more than once (repetition is not allowed).
5.2 Find the probability P(E or F) if E and F are mutually exclusive, P(E)=0.18, and P(F)=0.33.
Two events are disjoint if they have no outcomes in common. Another name for disjoint events is mutually exclusive events. To find the probability P(E or F), use the general addition rule shown below. P(A or B) = P(A)+P(B)−P(A and B) First determine the probability P(E and F). Recall that E and F are mutually exclusive. P(E and F)=0 Find the probability P(E or F). P(E or F)=P(E)+P(F)−P(E and F) P(E or F)=0.18+0.33−0 P(E or F)=0.51
5.4 The probability that a randomly selected individual in a country earns more than $75,000 per year is 12.5%. The probability that a randomly selected individual in the country earns more than $75,000 per year, given that the individual has earned a bachelor's degree, is 12.5%. Are the events "earn more than $75,000 per year" and "earned a bachelor's degree" independent?
Two events are independent if the occurrence of event E in a probability experiment does not affect the probability of event F. Independence can be expressed using conditional probabilities. The conditional probability P(F E) is the probability that the event F occurs, given that the event E has occurred. The notation P(F E) is read "the probability of event F given event E." Two events E and F are independent if P(E F)=P(E) or, equivalently, if P(F E) = P(F). The probability that a randomly selected individual in the country earns more than $75,000 per year, given that the individual has earned a bachelor's degree, and the probability that a randomly selected individual in a country earns more than $75,000 per year are the same and equal to 12.5%. Thus, the events "earn more than $75,000 per year" and "earned a bachelor's degree" are independent.
5.2 A golf ball is selected at random from a golf bag. If the golf bag contains 3 violet balls, 2 fuchsia balls, and 9 teal balls, find the probability of the following event. The golf ball is violet or fuchsia.
Use the addition rule for disjoint events to find the probability. The rule states that if E and F are disjoint (or mutually exclusive) events, then P(E or F)=P(E)+P(F). Applying the addition rule to this problem gives the following. P(violet or fuchsia)=P(violet)+ P(fuchsia) To find the probabilities of drawing a violet or fuchsia ball, first determine the size of the sample space S. n(S)=14 Now identify, N(violet), the number of violet balls in the golf bag. N(violet)=3 Find the probability of drawing a violet ball, rounding to three decimal places. P(violet) = n(violet) / n(S) = 3 / 14 = 0.214 Now identify, N(fuchsia), the number of fuchsia balls in the golf bag. N(fuchsia)=2 Next find the probability of drawing a fuchsia ball, rounding to three decimal places. P(fuchsia) = n(fuchsia) / n(S) = 2 / 14 = 0.143 Add the two probabilities to find the probability that a randomly selected golf ball is violet or fuchsia. P(violet or fuchsia) = 0.214+0.143 = 0.357 Therefore, the probability of selecting a violet or fuchsia golf ball from the golf bag is 0.357.
5.3 Suppose that events E and F are independent, P(E)=0.2, and P(F)=0.3. What is the P(E and F)?
Use the multiplication rule for independent events to find the probability. If E and F are independent events, then the probability P(E and F) is given by the formula below. P(E and F)=P(E)•P(F) The probability P(E and F) is 0.2•0.3=0.06.
5.4 Suppose there is a 11.8% probability that a randomly selected person aged 20 years or older is a jogger. In addition, there is a 20.6% probability that a randomly selected person aged 20 years or older is male, given that he or she jogs. What is the probability that a randomly selected person aged 20 years or older is male and jogs? Would it be unusual to randomly select a person aged 20 years or older who is male and jogs? The probability that a randomly selected person aged 20 years or older is male and jogs is _____ (Round to three decimal places as needed.) Would it be unusual? yes and no
a) 0.024 b) yes
5.4 The following data represent the number of drivers involved in a fatal crash in 2016 in various light and weather conditions. Complete parts (a) through (c) below. Click the icon to view the data for fatal crashes in 2016. (a) Among fatal crashes in snow/sleet, what is the probability that a randomly selected fatal crash occurs when it is dark (without light)? The probability that a randomly selected fatal crash in snow/sleet occurs when it is dark (without light) is approximately _____ (Round to three decimal places as needed.) (b) Among fatal crashes when it is dark (without light), what is the probability that a randomly selected fatal crash occurs in snow/sleet? The probability that a randomly selected fatal crash when it is dark (without light) occurs in snow/sleet is approximately _____. (Round to three decimal places as needed.) (c) Is dark, but lighted more dangerous in normal weather or in rain? Explain. Choose the correct choice below. A. Dark, but lighted is more dangerous in rain because the conditional probability of a fatal crash in rain is less than that of a fatal crash in normal weather. B. Dark, but lighted is more dangerous in rain because the conditional probability of a fatal crash in rain is greater than that of a fatal crash in normal weather. C. Dark, but lighted is more dangerous in normal weather because the conditional probability of a fatal crash in normal weather is less than that of a fatal crash in rain. D. Dark, but lighted is more dangerous in normal weather because the conditional probability of a fatal crash in normal weather is greater than that of a fatal crash in rain.
a) 0.351 b) 0.016 c)B. Dark, but lighted is more dangerous in rain because the conditional probability of a fatal crash in rain is greater than that of a fatal crash in normal weather. Note: It can be determined in what weather it is more dangerous by comparing two conditional probabilities. The larger probability indicates that a fatal crash is more likely.
5.3 Determine whether the events E and F are independent or dependent. Justify your answer. (a) E: A person living at least 70 years. F: The same person regularly handling venomous snakes. A. E and F are independent because living at least 70 years has no effect on the probability of a person regularly handling venomous snakes. B. E and F are independent because regularly handling venomous snakes has no effect on the probability of a person living at least 70 years. C. E and F are dependent because regularly handling venomous snakes can affect the probability of a person living at least 70 years. D. E and F are dependent because living at least 70 years has no effect on the probability of a person regularly handling venomous snakes. (b) E: A randomly selected person at school A having a high GPA. F: A randomly selected person at school B having a low GPA. A. E can affect the probability of F because the people were randomly selected, so the events are dependent. B. E cannot affect F and vice versa because the people were randomly selected, so the events are independent. C. E can affect the probability of F, even if the two people are randomly selected, so the events are dependent. D. E cannot affect F because "person 1 at school A having a high GPA" could never occur, so the events are neither dependent nor independent. (c) E: The consumer demand for synthetic diamonds. F: The amount of research funding for diamond synthesis. A. The consumer demand for synthetic diamonds could not affect the amount of research funding for diamond synthesis, so E and F are independent. B. The consumer demand for synthetic diamonds could affect the amount of research funding for diamond synthesis, so E and F are dependent. C. The amount of research funding for diamond synthesis could affect the consumer demand for synthetic diamonds, so E and F are dependent.
a) C. E and F are dependent because regularly handling venomous snakes can affect the probability of a person living at least 70 years. b) B. E cannot affect F and vice versa because the people were randomly selected, so the events are independent. c) B. The consumer demand for synthetic diamonds could affect the amount of research funding for diamond synthesis, so E and F are dependent.
5.3 Determine whether the events E and F are independent or dependent. Justify your answer. (a) E: A person having an at-fault accident. F: The same person being prone to road rage. A. E and F are independent because being prone to road rage has no effect on the probability of a person having an at-fault accident. B. E and F are dependent because having an at-fault accident has no effect on the probability of a person being prone to road rage. C. E and F are independent because having an at-fault accident has no effect on the probability of a person being prone to road rage. D. E and F are dependent because being prone to road rage can affect the probability of a person having an at-fault accident. (b) E: A randomly selected person finding cheese revolting. F: A different randomly selected person finding cheese delicious. A. E cannot affect F because "person 1 finding cheese revolting" could never occur, so the events are neither dependent nor independent. B. E cannot affect F and vice versa because the people were randomly selected, so the events are independent. C. E can affect the probability of F, even if the two people are randomly selected, so the events are dependent. D. E can affect the probability of F because the people were randomly selected, so the events are dependent. (c) E: The unusually foggy weather in London on May 8. F: The number of car accidents in London on May 8. A. The unusually foggy weather in London on May 8 could affect the number of car accidents in London on May 8, so E and F are dependent. B. The number of car accidents in London on May 8 could affect the unusually foggy weather in London on May 8, so E and F are dependent. C. The unusually foggy weather in London on May 8 could not affect the number of car accidents in London on May 8, so E and F are independent.
a) D. E and F are dependent because being prone to road rage can affect the probability of a person having an at-fault accident. b)B. E cannot affect F and vice versa because the people were randomly selected, so the events are independent. c)A. The unusually foggy weather in London on May 8 could affect the number of car accidents in London on May 8, so E and F are dependent.
5.1 Determine if the following statement is true or false. Probability is a measure of the likelihood of a random phenomenon or chance behavior. Choose the correct answer below. a. True b. False
a. True
5.3 Determine if the following statement is true or false. When two events are disjoint, they are also independent. Choose the correct answer below. a. True b. False
b. False Note: The correct answer is False because two events are disjoint if they have no outcomes in common. In other words, the events are disjoint if, knowing that one of the events occurs, we know the other event did not occur. Independence means that one event occurring does not affect the probability of the other event occurring. Therefore, knowing two events are disjoint means that the events are not independent.
5.5 Fill in the blank. A ________ is an arrangement of r objects chosen from n distinct objects without repetition and without regard to order.
combination Note: If n≥0 is an integer, the factorial symbol, n!, is defined as follows. This is not an arrangement of objects, but instead is a mathematical symbol. n! = n(n−1)•••••3•2•1 0! = 1 1!=1 Conditional probability is represented by the notation P(F|E), which is read "the probability of event F given event E." It is the probability that the event F occurs, given that the event E has occurred. Conditional probability is a type of probability, not an arrangement of objects.. A combination is a collection, without regard to order, of n distinct objects without repetition. The symbol nCr represents the number of combinations of n distinct objects taken r at a time.
5.1 Fill in the blank. In probability, a(n) ________ is any process that can be repeated in which the results are uncertain.
experiment Note: An event is any collection of outcomes from a probability experiment. An unusual event is an event that has a low probability of occurring. The sample space, S, of a probability experiment is the collection of all possible outcomes. In probability, an experiment is any process with uncertain results that can be repeated. The result of any single trial of the experiment is not known ahead of time. However, the results of the experiment over many trials produce regular patterns that enable one to predict with remarkable accuracy.
5.3 Fill in the blank. Two events E and F are ________ if the occurrence of event E in a probability experiment does not affect the probability of event F.
independent Note: Two events E and F are independent if the occurrence of event E in a probability experiment does not affect the probability of event F.
5.3 Fill in the blank. The word and in probability implies that we use the ________ rule.
multiplication Note: The multiplication rule states that if events E, F, G, ... are independent, then P(E and F and G and ...)=P(E)•P(F)•P(G) .... Accordingly, and probabilities use the multiplication rule. The general addition rule states that for any two events E and F, P(E or F)=P(E)+F(F)−P(E and F). Accordingly, or probabilities use the addition rule. There is no subtraction rule in computing probabilities. The general addition rule states that for any two events E and F, P(E or F)=P(E)+F(F)−P(E and F). The multiplication rule states that if events E, F, G, ... are independent, then P(E and F and G and ...)=P(E)•P(F)•P(G) ....
5.4 The following data represent the number of drivers involved in a fatal crash in 2016 in various light and weather conditions. Complete parts (a) through (c) below. Click the icon to view the data for fatal crashes in 2016.
(a) Among fatal crashes in unknown weather conditions, what is the probability that a randomly selected fatal crash occurs when it is daylight? The probability that an event F occurs, given that the event E has occurred is called a conditional probability and is written P(F|E). E is the event which is known to have occurred. It is known that the fatal crash occurred in unknown weather conditions. F is the event for which the probability is sought. The question asks for the probability that the fatal crash occurs when it is daylight. If E and F are any two events, then P(F|E)=P(E and F) / P(E)=N(E and F) / N(E). The probability of event F occurring, given the occurrence of event E, is found by dividing the probability of E and F by the probability of E, or dividing the number of outcomes in E and F by the number of outcomes in E. The probability of event E and F is the probability the fatal crash occurs in unknown weather conditions and when it is daylight. Since the data is given as number of occurrences, use that form of the definition. Therefore, the probability that a randomly selected fatal crash in unknown weather conditions occurs when it is daylight can be found as shown. P(F|E)=N(E and F) / N(E) = fatal crashes in unknown weather conditions and occur when it is daylight / fatal crashes in unknown weather conditions From the table, there were 810 fatal crashes that occurred both in unknown weather conditions and when it was daylight. The number of fatal crashes that occurred in unknown weather conditions is the sum of all the entries in the indicated row. 810+255+548+71+133=1817 Apply the definition and divide to find the probability, rounding to three decimal places. P(F|E) = 810 / 1817 ≈ 0.446 Therefore, the probability that a randomly selected fatal crash in unknown weather conditions occurs when it is daylight is approximately 0.446. (b) Among fatal crashes when it is daylight, what is the probability that a randomly selected fatal crash occurs in unknown weather? Notice that for the second question, the event that is known to have occurred is interchanged with the event whose probability is being sought. For this question, it is known that the fatal crash occurred when it was daylight. Since the two events involved are the same, the combined event E and F is the same. P(F|E)=N(E and F) / N(E) = fatal crashes in unknown weather conditions and occur when it is daylight / fatal crashes that occur when it is daylight There were 810 fatal crashes that occurred in unknown weather conditions and when it was daylight. The number of fatal crashes occurred when it was daylight is the sum of all the entries in the indicated column. 14,307+875+219+125+810=16,336 Apply the definition and divide to find the probability, rounding to three decimal places. P(F|E) = 810 / 16,336 ≈ 0.050 Therefore, the probability that a randomly selected fatal crash in daylight occurs in unknown weather conditions is approximately 0.050. (c) Is the daylight more dangerous in normal weather or in snow/sleet? Explain. It can be determined whether it is more dangerous in normal weather or in snow/sleet by comparing two conditional probabilities. Let A be the event that a fatal crash occurs in daylight, B be the event that a fatal crash occurs in normal weather, and C be the event that a fatal crash occurs in snow/sleet. Comparing P(A|B) and P(A|C) will determine the weather conditions that make driving in daylight more dangerous. Start by computing P(A|B), rounding to three decimal places. Recall that P(A|B)=N(A and B)N(B). P(A|B) = 14,307 / 29,581 ≈ 0.484 Then compute P(A|C), rounding to three decimal places. P(A|C) = 219 / 445 ≈ 0.492 Compare the two probabilities. The daylight is more dangerous in snow/sleet because the conditional probability of a fatal crash in snow/sleet is greater than that of a fatal crash in normal weather.
5.5 Fill in the blanks. The factorial symbol, n!, is defined as n!=_______ and 0!=_______.
n!=n(n−1)•...•3•2•1, 1 Note: If n≥0 is an integer, the factorial symbol, n!, is defined by the formulas below. n!=n(n−1)•...•3•2•1 1!=1 0!=1
