MATH-164 - Chapter 5
A bag of 100 tulip bulbs purchased from a nursery contains 35 red tulip bulbs, 30 yellow tulip bulbs, and 35 purple tulip bulbs. (a) What is the probability that a randomly selected tulip bulb is red? (b) What is the probability that a randomly selected tulip bulb is purple? (c) Interpret these two probabilities.
(a) 0.35 (b) 0.35 (c) If 100 tulip bulbs were sampled with replacement, one would expect about 35 of the bulbs to be red and about 35 of the bulbs to be purple. (the word "about" not the word "exactly")
Find the value of the combination. 11C2
11C2=55 (Statcrunch/data/compute, expression, comb(11,2))
Find the value of the permutation. 2P1
2P1=2 (Statcrunch/data/compute, expression, Perm(2,1))
Find the value of the factorial. 7!
7!=5040 (7*6*5*4*3*2*1=5040)
On the basis of a survey of 1000 families with six children, the probability of a family having six girls is 0.0054. A.Empirical method B.Classical method C.Subjective method D.It is impossible to determine which method is used.
A.Empirical method
Which type of compound event is generally associated with multiplication? Which is generally associated with addition?
An ''AND' compound event is generally associated with multiplication; an ''OR' compound event is generally associated with addition.
What does it mean for an event to be unusual? Why should the cutoff for identifying unusual events not always be 0.05?
An event is unusual if it has a low probability of occurring. The choice of a cutoff should consider the context of the problem.
Subjective Probability Assessment
An opinion or judgment by a decision maker about the likelihood of an event. is a probability that is determined based on personal judgment.
Explain the Law of Large Numbers. How does this law apply to gambling casinos?
As the number of repetitions of a probability experiment increases, the proportion with which a certain outcome is observed gets closer to the probability of the outcome. This applies to casinos because they are able to make a profit in the long run because they have a small statistical advantage in each game.
Determine whether the probabilities below are computed using the classical method, empirical method, or subjective method.The probability of having six girls in an six-child family is 0.015625. A.Empirical method B.Classical method C.Subjective method D.It is impossible to determine which method is used.
B.Classical method
Brad and Allison have three girls. Brad tells Allison that he would like one more child because they are due to have a boy. What do you think of Brad's logic?
Brad is incorrect due to the nonexistent Law of Averages. The fact that Brad and Allison had three girls in a row does not matter. The likelihood the next child will be a boy is about 0.5.
What method of assigning probabilities to a simple event uses relative frequencies?
Empirical
(d) On the basis of clinical trials, the probability of efficacy of a new drug is 0.84.
Empirical method
In probability, a(n) ________ is any process that can be repeated in which the results are uncertain.
Experiment
The notation P(F E)means the probability of event ____ given event ___. .
F, E
Suppose that events E and F are independent, P(E)=0.3, and P(F)=0.7. What is the P(E and F)?
If E and F are independent, Then Here P(E) = 0.3 P(F) =0.7 Thus P(E and F ) = 0.3 *0.7 =0.21 The probability is 0.21.
If E and F are not disjoint events, then P(E or F)=________.
If E and F are not disjoint events, then P(E or F)= P(E) + P(F) - P(E and F)
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
Suppose that P(E)=0.7, P(F)=0.5, and P(E and F)=0.24. Are events E and F independent? Why?
No, because P(E and F)≠P(E)•P(F).
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)?
P(E and F)=0.56 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 0.7*0.8=0.56
If E and F are disjoint events, then P(E or F) =
P(E) + P(F).
If E and F are disjoint events, then P(E or F)=
P(E)+P(F)
P(E)=relative frequency of E
P(E)=frequency of E/number of trials of experiment
Suppose that E and F are two events and that N(E and F)=460 and N(E)=800. What is P(F|E)?
P(F|E)≈0.575 (Round to three decimal places asneeded.) Apply the definition for P(F|E) using the number of outcomes. P(F|E)=N(E and F)N(E)=460/800 Then divide to find the probability. 460/800≈0.575
_________________________is a technique used to recreate a random event.
Simulation
(c) According to a sports analyst, the probability that a football team will win the next game is 0.36.
Subjective method
Suppose 60 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 60 cars is 205320. (Statcrunch/data/compute, expression, Perm(60,3))
In a certain card game, the probability that a player is dealt a particular hand is 0.36. Explain what this probability means. If you play this card game 100 times, will you be dealt this hand exactly 36 times? Why or why not
The probability 0.36 means that approximately 36 out of every 100 dealt hands will be that particular hand. No, you will not be dealt this hand exactly 36 times since the probability refers to what is expected in the long-term, not short-term.
Suppose that events E and F are independent, P(E)=0.6, and P(F)=0.7. What is the P(E and F)?
The probability P(E and F) is 0.42 P(E and F)=P(0.6)*P(0.7)=0.42
Find the probability P(Ec)if P(E)=0.43.
The probability P(Ec)is 0.57. (Simplify your answer.) P(Ec)=1−P(E) P(Ec)=1-.43=.57
A woman has eight skirts and six blouses. Assuming that they all match, how many different skirt-and-blouse combinations can she wear?
The woman can wear 48 different skirt-and-blouse combinations. p*q different ways (8*6=48)
Why is the following not a probability model? Color Probability Red 0.2 Green −0.4 Blue 0.1 Brown 0.4 Yellow 0.2 Orange 0.5 Determine why it is not a probability model. Choose the correct answer below.
This is not a probability model because at least one probability is less than 0.
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?
Yes
A Deck of Cards A standard deck of cards contains 52 cards, as shown in Figure 9. One card is randomly selected from the deck. (a) Compute the probability of randomly selecting a two or three from a deck of cards. (b) Compute the probability of randomly selecting a two or three or four from a deck of cards. (c) Compute the probability of randomly selecting a two or club from a deck of cards.
a) Standard deck contains 4 two's and 4 three's P(Two OR three) = 4/52 + 4/52 = 0.1538 b) Standard deck contains 4 two's and 4 three's and 4 four's P(Two OR three OR four) = 4/52 + 4/52 + 4/52 = 0.2308 c) P(Two OR club) = P(two) + P(Club) - P(two of club) = 4/52 + 13/52 - 1/52 = 0.3077
A grocery bag can be classified as paper or plastic. 91% of grocery bags are classified as plastic. (a) What is the probability that two randomly selected grocery bags are plastic? (b) What is the probability that seven randomly selected grocery bags are plastic? (c) What is the probability that at least one of seven randomly selected grocery bags are paper? Would it be unusual or usual that at least one of the seven randomly selected grocery bags is paper?
a. (.91)^2=.8281 b. (.91)^7=.5168 c. 1 - prob that none of the 7 are paper (i.e. All are plastic) 1 - (.91)^7 = .4832 It would be usual
The probability that a randomly selected 5-year-old male garter snake will live to be 6 years old is 0.95792. (a) What is the probability that two randomly selected 5-year-old male garter snakes will live to be 6 years old? (b) What is the probability that six randomly selected 5-year-old male garter snakes will live to be 6 years old?
a. (a) The probability that two randomly selected 5-year-old male garter snakes will live to be 6 years old is 0.91761 (0.95792)^2 (b) The probability that six randomly selected 5-5-year-old male garter snakes will live to be 6 years old is 0.77264 (0.95792)^6
Suppose Aaron is going to build a playlist that contains 13 songs. In how many ways can Aaron arrange the 13 songs on the playlist?
can arrange the 13 songs on the playlist in 6227020800 different ways. (13*12*11*10*9*8*7*6*5*4*3*2*1=5040)
A(n) __________ is any collection of outcomes from a probability experiment.
event
disjoint event
events that have no outcomes in common
disjoint events
events that have no outcomes in common. if they have no outcomes in common, that is, if knowing that one of the events occurs, we know that the other event did not occur.
combination
is a collection, without regard to order, in which r objects are chosen from n distinct objects with r≤n without repetition. The symbol nCr represents the number of combinations of n distinct objects taken r at a time.
event
is any collection of outcomes from a probability experiment. An event consists of one or more outcomes. We denote events with one outcome, sometimes called simple events, as ei. In general, events are denoted using capital letters such as E
Probability Model/Distribution
lists the possible outcomes of a probability experiment and each outcome's probability. A probability model must satisfy Rules 1 and 2 of the rules of probabilities. 1. The probability of any event E,P(E), must be greater than or equal to 0 and less than or equal to 1. That is, 0≤P(E)≤1. 2. The sum of the probabilities of all outcomes must equal 1. That is, if the sample space S={e1,e2,⋯,en}, then P(e1)+P(e2)+⋯+P(en)=1
independence
means that one event occurring does not affect the probability of the other event occurring.
outcome
outcome = one trial
Select the correct choice that completes the sentence below. The word ___________ suggests an unpredictable result or outcome.
random
A________________represents scenarios where the outcome of any particular trial of an experiment is unknown, but the proportion (or relative frequency) a particular outcome is observed approaches a specific value.
random process
Sample Space (Probability)
the set of all possible outcomes (S) sample space=all possible results
The law of large numbers
theory that says if you want to predict how likely an event is to occur, you will get the most accurate answer by looking at the largest number of cases where it might. As the number of repetitions of a probability experiment increases, the proportion with which a certain outcome is observed gets closer to the probability of the outcome.
According to a center for disease control, the probability that a randomly selected person has hearing problems is 0.141. The probability that a randomly selected person has vision problems is 0.087. 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?
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.
A survey of 500 randomly selected high school students determined that 243 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 0.486. (243/500=0.486) (b) Choose the correct answer below. If 1,000 high school students were sampled, it would be expected that about 486 of them play organized sports. (about not exactly)
A bag of 30 tulip bulbs contains 12 red tulip bulbs, 9 yellow tulip bulbs, and 9 purple tulip bulbs. Suppose two tulip bulbs are randomly selected without replacement from the bag. (a) What is the probability that the two randomly selected tulip bulbs are both red? (b) What is the probability that the first bulb selected is red and the second yellow? (c) What is the probability that the first bulb selected is yellow and the second red? (d) What is the probability that one bulb is red and the other yellow?
(a) The probability that both bulbs are red is 0.152. (Round to three decimal places as needed.) 12/30=0.4 11/29=0.379 0.4*0.379=0.152 (b) The probability that the first bulb is red and the second is yellow is 0.124 (Round to three decimal places as needed.) 12/30=0.4 9/29=0.310 0.4*0.310=0.124 (c) The probability that the first bulb is yellow and the second is red is 0.124 (Round to three decimal places as needed.) 9/30=0.3 12/29=0.414 0.3*0.414=0.124 (d) The probability that one bulb is red and one is yellow is 0.248 (Round to three decimal places as needed.) 0.124+0.124=0.248
For the fiscal year 2007, a tax authority audited 1.82% of individual tax returns with income of $100,000 or more. Suppose this percentage stays the same for the current tax year. What is the probability that two randomly selected returns with income of $100,000 or more will be audited?
The probability is 0.000331 P(E and F)=P(E)•P(F) 0.0182^2=0.00033124
For the month of January in a certain city, 76% of the days are cloudy. Also in the month of January in the same city, 32% of the days are cloudy and snowy. What is the probability that a randomly selected day in January will be snowy if it is cloudy?
The probability is approximately 0.421. (Round to three decimal places as needed.) (0.32/0.76=0.421)
Suppose there is a 12.8% probability that a randomly selected person aged 25 years or older is a smoker. In addition, there is a 28.9% probability that a randomly selected person aged 25 years or older is female, given that he or she smokes. What is the probability that a randomly selected person aged 25 years or older is female and smokes? Would it be unusual to randomly select a person aged 25 years or older who is female and smokes?
The probability that a randomly selected person aged 25 years or older is female and smokes is 0.037 (Round to three decimal places as needed.). (0.128*.289=0.037) Would it be unusual? Yes (if its under 5% its unusual, so in this case it is)
Suppose you just received a shipment of six televisions. Four of the televisions are defective. If two televisions are randomly selected, compute the probability that both televisions work. What is the probability at least one of the two televisions does not work?
The probability that both televisions work is 067. 6 TV /4 Bad/2 working divide working tv by the number of TV, then multiply by the remaining working TV and the remaining available TV 2/6*1/5=0.067 The probability that at least one of the two televisions does not work is 0.933 1-0.067=0.933
A golf ball is selected at random from a golf bag. If the golf bag contains 6 brown balls, 5 green balls, and 9 yellow balls, find the probability of the following event. The golf ball is brown or green.
The probability that the golf ball is brown or green is 0.550 (Type an integer or a decimal rounded to three decimal places as needed.) 6+5+9=20 6/20=0.3 5/20=0.25 0.3+0.25=0.50
Find the probability P(E or F) if E and F are mutually exclusive, P(E)=0.25,and P(F)=0.51.
The probability P(E or F) is 0.76. (Simplify your answer.) P(E or F)=P(E)+P(F)−P(E and F) P(E or F)=.25+.51-0=0.76
Find the probability P(E or F) if E and F are mutually exclusive, P(E)=0.34, and P(F)=0.51.
The probability P(E or F) is 0.85. P(E and F)=0 P(E or F)=P(E)+P(F)−P(E and F) P(E or F)=0.34+0.51−0 P(E or F)=0.85
Determine whether the events E and F are independent or dependent. Justify your answer (a) E: A person attaining a position as a professor. F: The same person attaining a PhD. (b) E: A randomly selected person planting tulip bulbs in October. F: A different randomly selected person planting tulip bulbs in April. (c) E: The war in a major oil-exporting country. F: The price of gasoline.
a. E and F are dependent because attaining a PhD can affect the probability of a person attaining a position as a professor. b. E cannot affect F and vice versa because the people were randomly selected, so the events are independent. c. The war in a major oil-exporting country could affect the price of gasoline, so E and F are dependent.
In a recent poll, a random sample of adults in some country (18 years and older) was asked, "When you see an ad emphasizing that a product is "Made in our country," are you more likely to buy it, less likely to buy it, or neither more nor less likely to buy it?" The results of the survey, by age group, are presented in the following contingency table. Complete parts (a) through (c). Purchase_likelihood 18-34 35-44 45-54 55+ Total More likely 207 383 394 406 1390 Less likely 24 8 26 13 71 Neither more nor less likely 298 207 150 130 785 Total 529 598 570 549 2246 (a) What is the probability that a randomly selected individual is at least 55 years of age, given the individual is more likely to buy a product emphasized as "Made in our country"? (b) What is the probability that a randomly selected individual is more likely to buy a product emphasized as "Made in our country," given the individual is at least 55 years of age? (c) Are 18- to 34-year-olds more likely to buy a product emphasized as "Made in our country" than individuals in general?
a. The probability is approximately 0.292 (406/1390=0.292) b. The probability is approximately 0.740 (406/549=0.740) c. No, less likely (Compare the data from the last column.
unusual event
an event with a low probability of occurrence. Typically, an event with a probability less than 0.05 (or 5%) is considered unusual, but this cutoff point is not set in stone. The researcher and the context of the problem determine the probability that separates unusual events from not so unusual events. If an event does not show 51% of the likelihood it is unusual even if it is the largest individual number.
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 allow accurate predictions.
Probability
is the measure of the likelihood of a random phenomenon or chance behavior occurring.
The grade appeal process at a university requires that a jury be structured by selecting six individuals randomly from a pool of nine students and nine faculty. (a) What is the probability of selecting a jury of all students? (b) What is the probability of selecting a jury of all faculty? (c) What is the probability of selecting a jury of three students and three faculty?
total people = 9 + 9 = 18 a) P(selecting a jury of all students) = 9C6 / 18C6 = 84 / 18564 = 0.0045 b) P(selecting a jury of all faculty) = 9C6 / 18C6 = 84 / 18564 = 0.0045 c) P(selecting a jury of 3 students and 3 faculty) = 9C3 * 9C3 / 18C6 = 84 / 18564 = 0.3801
The data on the right represent the number of live multiple-delivery births (three or more babies) in a particular year for women 15 to 54 years old. Use the data to complete parts (a) through (d) below Age Number of Multiple Births 15-19 91 20-24 508 25-29 1639 30-34 2837 35-39 1851 40-44 373 45-54 119
(a) Determine the probability that a randomly selected multiple birth for women 15-54 years old involved a mother 30 to 39 years old. P(30 to 39)=0.632 solution: total=91+508+1639+2837+1851+373+119=7418 N(S)=7418 (a)Out of 7418 women, 2837+1851 = 4688 mother are 30 to 39 years old so the probability that a randomly selected multiple birth for women 15-54 years old involved a mother 30 to 39 years old is P(involved a mother 30 to 39 years old) = 4688 / 7418= 0.632 b)P(not involved a mother 30 to 39 years old) =1- P(involved a mother 30 to 39 years old) = 1 - 0.632 = 0.368 c)Out of 73418 women, 7418-119= 7299 mother are less than 45 years old so the probability that a randomly selected multiple birth for women 15-54 years old involved a mother less than 45 years old is P(involved a mother less than 45 years old) = 7299 / 7418 = 0.984 d)Out of 7418 women, 373+119 = 492 mother are at least 40 years old so the probability that a randomly selected multiple birth for women 15-54 years old involved a mother at least 40 years old is P(involved a mother at least 40 years old) = 492 / 7418 = 0.066 Interpret this result. Select the correct choice below and fill in the answer box to complete your choice. If 1000 multiple births for women 15-54 years old were randomly selected, we would expect about 66 of them to involve a mother who was at least 40 years old. Is a multiple birth involving a mother who was at least 40 years old unusual? No, because the probability of a multiple birth involving a mother who was at least 40 years old is greater than 0.05.
Suppose that two cards are randomly selected from a standard 52-card deck. (a) What is the probability that the first card is a club and the second card is a club if the sampling is done without replacement? (b) What is the probability that the first card is a club and the second card is a club if the sampling is done with replacement?
(a) If the sampling is done without replacement, the probability that the first card is a club and the second card is a club is 0.059 (Round to three decimal places as needed.) 13/52*12/51=156/2652=0.5882=0.059 (b) If the sampling is done with replacement, the probability that the first card is a club and the second card is a club is 0.063 (Round to three decimal places as needed.) 13/52*13/52=169/2704=0.0625
Suppose that a computer chip company has just shipped 5,000 computer chips to a computer company. Unfortunately, 20 of the chips are defective. (a) Compute the probability that two randomly selected chips are defective using conditional probability. (b) The probability that the first randomly selected chip is defective is 205,000=0.004=0.4%. Compute the probability that two randomly selected chips are defective under the assumption of independent events.
(a) The probability is 0.00001520. 20*19=380 then 5,000*4,999=24,995,000 380/24,995,000=0.00001520304 (b) The probability is 0.00001600 (Round to eight decimal places as needed.) 0.004*0.004=0.00001600 When small samples are taken from large populations without replacement, the assumption of independence does not significantly affect the probability. Based on the results, what does this mean? The probabilities are nearly the same.
Suppose you just purchased a digital music player and have put 11 tracks on it. After listening to them you decide that you like 4 of the songs. With the random feature on your player, each of the 11 songs is played once in random order. Find the probability that among the first two songs played (a) You like both of them. Would this be unusual? (b) You like neither of them. (c) You like exactly one of them. (d) Redo (a)-(c) if a song can be replayed before all 11 songs are played.
(a) The probablility that you like both songs is 0.109 (4/11*3/10=12/110=0.109) Would it be unusual for you to like both of the songs? No (b) The probability that you like neither song is 0.382 (7/11*6/10=42/110=0.382) (c) The probability that you like exactly one song is 0.509 (1-0.109-0.382=0.509) (d) The probability that you like both songs is 0.132 (4/11*4/11=16/121=0.132) The probability that you like neither song is 0.405. (7/11*7/11=49/121=0.405) The probability that you like exactly one song is 0.463 (1-0.132-0.405=0.463)
Find the probability of the indicated event if P(E)=0.25 and P(F)=0.40. Find P(E or F) if P(E and F)=0.10.
P(E or F)=0.55 (Simplify your answer.) P(E or F)=P(E)+P(F)−P(E and F). P(E or F)=.25+.40-.10=0.55
