PROBABILITY
If the letters k, n, i, c, k, k, n, a, c, and k are chosen at random, what is the probability that they spell the word "knickknack"?
1 / 37800
A die is rolled 8 times. Find the probability of rolling the following. Exactly 8 sixes.
The probability is 5.954 × 10^−7 (1/6)^8
A certain website requires users to log on using a security password. (a) If passwords must consist of six letters followed by a single digit, determine the total number of possible distinct passwords. (b) If passwords must consist of six non-repetitive letters followed by a single digit, determine the total number of possible distinct passwords.
(A) 3,089,157,760 (B) 1,657,656,000
For the experiment described below, let x determine a random variable, and use your knowledge of probability to prepare a probability distribution. Three balls are drawn (with replacement) from a bag that contains 4 green balls and 6 yellow balls. The number of yellow balls is counted.
.0640 .2880 .4320 .2160
Two cards are drawn without replacement from an ordinary deck. Find the probability that two aces are drawn.
1/221
A certain species of animal has a 1/2 probability of any one offspring being male and a 1/2 probability of any one offspring being female. Fill in the remaining probabilities on the tree diagram and use that information to find the probability that an adult female of the species produces three offspring that are all females, given that the second is a female.
1/8 for all the branches The probability that an adult female of the species produces three offspring that are all females, given that the second is a female, is 1/4
A shipment of 12 printers contains 2 that are defective. Find the probability that a sample of size 4, drawn from the 12, will not contain a defective printer.
12 - 2 = 10 non deffective C(10,4) =210 C(12,4) = 495 210/495 = 14/33
Find the expected value of the random variable. #2
2(0.1)+3(0.3)+4(0.3)+5(0.4) = 4
A couple has narrowed down the choices of a name for their new baby to 6 first names and 3 middle names. How many different first- and middle-name arrangements are possible?
6 * 3 = 18
Find the future value of the ordinary annuity. Interest is compounded annually. R=$9000; i=0.10; n=5
9000[ (1+0.10)^5 - 1 / 0.10 ] The future value of the ordinary annuity is $54945.9
What is a combination?
A combination is a group of items taken without regard to their order.
In how many ways can a hand of 10 hearts be chosen from an ordinary deck? Formula: C(n,r) = n! / (n - r)!r! So 13! / (13 - 10)!10! = 286
A hand of 10 hearts can be chosen in nothing 286 ways.
For the experiment described below, let x determine a random variable, and use your knowledge of probability to prepare a probability distribution. Three balls are drawn (with replacement) from a bag that contains 6 blue balls and 5 yellow balls. The number of yellow balls is counted
A probability distribution for the given experiment is as follows. x0123P(x) .1623 .4057 .3381 .0939
A die is tossed. What is the probability of rolling a number greater than 4? The sample space is S={1, 2, 3, 4, 5, 6}. E is the event the die rolled is greater than 4. E={5, 6} P(E)=n(E)n(S)=2/6= 1/3
A single fair die is tossed. Find the probability of rolling a number greater than 5. What is the probability of rolling a number greater than 5? 1/6
Find the present value (the amount that should be invested now to accumulate the following amount) if the money is compounded as indicated. $9000 at 8% compounded semiannually for 3
A= 9000 r= 0.08 m=2 t=3 i=0.08/2 n= 2*3 or 6 P = A / (1+ i)^n P = 9000 / (1+0.08/2)^6 = 7112.830732 P= 7112.83
Evaluate C(c,(c−1)) =
C
A bag contains 3 red apples and 4 yellow apples. A sample of 3 apples is drawn. Find the probability that the sample contains more red than yellow apples.
C(3,2) * C(4,1) + C(3,3) * C(4,0) 3 4 1 1 C(7,3) = 35 3*4+1*1/35 = 13/35
Decide whether the exercise involves permutations or combinations, and then solve the problem. There are 8 rotten apples in a crate of 29 apples. (a) How many samples of 3 apples can be drawn from the crate? (b) How many samples of 3 could be drawn in which all 3 are rotten? (c) How many samples of 3 could be drawn in which there are two good apples and one rotten one?
Combinations PROBLEM A = 3654 B= 56 C = 1680
The coach of a softball team has 8 good hitters and 9 poor hitters. He chooses 5 hitters at random. Complete parts (a) through (c) below.
He can choose 4 good hitters and 1 poor hitter in 630 ways. He can choose 5 good hitters in 56 ways. He can choose at least 4 good hitters in 686 ways.
An accountant for a corporation forgot to pay the firm's income tax of $725,608.37 on time. The government charged a penalty of 7.5% interest for the 53 days the money was late. Find the total amount (tax and penalty) that was paid. Assume 365 days in a year.
P = 725608.37 | r = 7.5% = 0.075 | t= 53/365 A = P(1+rt) A = 725608.37(1+0.075*53/365) A = 733510.5433 The total amount paid was $733510.54
Find the expected payback for a game in which you bet $8 on any number from 0 to 399. If your number comes up, you get $500.
The expected payback is −6.75.
A random study reported that 70% of very-low-birth-weight babies graduate from high school by age 20. If 130 very-low-birth-weight babies are followed through high school, how many would you expect to graduate from high school?
The number of very-low-birth-weight babies expected to graduate is 91 .7*130 = 91
A recent survey found that 86% of first-year college students were involved in volunteer work at least occasionally. Suppose a random sample of 12 college students is taken. Find the probability that exactly 8 students volunteered at least occasionally.
The probability is 0.0569. this is how to do it https://www.chegg.com/homework-help/volunteering-recent-survey-found-83-first-year-college-stude-chapter-8.4-problem-59e-solution-9780321748676-exc
According to a recent report, 66.1% of men and 50.2% of women in the United States were overweight. Given that 48.7% of Americans are men and 51.3% are women, find the probability that a randomly selected American fits the following description. (a) An overweight man (b) Overweight (c) Are the events that an adult is a man and that an adult is overweight independent? Explain.
The probability that the adult is a man is 0.487 The probability that the adult is overweight, given that the adult is a man is 0.661 0.487*0.661 = 0.322 0.513 are women 0.502 were overweight 0.513*0.502= 0.258 0.322+0.258= 0.580 Observe P(o) = 0.580 and P (o|M) = 0.661 No. The events are not independent because the probability that an adult is overweight is not equal to the probability that an adult is overweight, given that the adult is a man. Therefore, the events that an adult is a man and that an adult is overweight are not independent.
A lottery game requires that you pick 5 different numbers from 1 to 66. If you pick all 5 winning numbers, you win the jackpot. If you pick 4 of the 5 numbers correctly, you win $300,000. In how many ways can you pick exactly 4 of the 5 winning numbers without regard to order?
There are 305mways to pick exactly 4 of the 5 winning numbers without regard to order. 5 x 61 = 305
Two cards are drawn without replacement from an ordinary deck, find the probability that the second is a king, given that the first is a king.
What is the conditional probability? 1/17
From a group of 6 women and 5 men, a delegation of 3 is selected. Find the expected number of women in the delegation.
What is the expected number of women? 1.64
hat is the probability of winning a lottery in which you must choose 6 numbers from the numbers 1 through 18?
What is the probability of winning the lottery? 1 / 18564 18! / 6!( 18 - 6 )! = 18564 how to do it: https://www.chegg.com/homework-help/questions-and-answers/probability-winning-lottery-must-choose-5-numbers-numbers-1-18-probability-winning-lottery-q29008239
Otitis media, or middle ear infection, is initially treated with an antibiotic. Researchers have compared two antibiotics, A and B, for their cost effectiveness. A is inexpensive, safe, and effective. B is also safe. However, it is considerably more expensive and it is generally more effective. Use the tree diagram to the right (where the costs are estimated as the total cost of medication, office visit, ear check, and hours of lost work) to answer the following. a. Find the expected cost of using each antibiotic to treat a middle ear infection. A tree diagram has a root that splits into 2 branches labeled A and B. Primary branch A splits into 2 secondary branches labeled 0.60 Cure 59.30 dollars and 0.40 No cure 96.15 dollars. Primary branch B splits into two secondary branches labeled 0.80 Cure 69.15 dollars and 0.20 No cure 106.00 dollars.0.600.400.800.20 b. To minimize the total expected cost, which antibiotic should be chosen?
a. The expected cost of using antibiotic A is 74.04. (Round to the nearest cent as needed.) b. The expected cost of using antibiotic B is 76.52 Antibiotic A minimizes the total expected cost. Therefore, antibiotic A should be chosen.
Pecan producers blow air through the pecans so that the lighter ones are blown out. The lighter-weight pecans are generally bad ("bad nuts") and the heavier ones tend to be better ("good nuts"). These "blow outs" and "non-blow outs" are often sold to tourists along the highway. Suppose that 50% of the pecan "blow outs" are good and 80% of the "non-blow outs" are good. a. If you purchase 50 pecans, what is the expected number of good nuts you will find if you purchase "blow outs?" b. If you purchase 50 pecans, what is the expected number of bad nuts you will find if you purchase "non-blow outs?"
a. The expected number of good nuts in 50 pecans from "blow outs" is 25. 50*.5=25 b. The expected number of bad nuts in 50 pecans from "non-blow outs" is 10. 1-.8 =.2 50*.2=10
One card is drawn from an ordinary deck of 52 cards. Find the probabilities of drawing the following cards. a. A 2 or 10 b. A red card or a 9 c. A 4 or a black 6 d. A diamond or a black card e. A face card or a
aLL THE ANSWER ARE THE SAME EVEN THO DIFFERENT NUMBERS https://www.chegg.com/homework-help/questions-and-answers/one-card-drawn-ordinary-deck-52-cards-find-probabilities-drawing-following-cards--4-7-b-re-q37324009
Find the future value of an ordinary annuity if payments are made in the amount R and interest is compounded as given. Then determine how much of this value is from contributions and how much is from interest. R=15,000; 4.1% interest compounded quarterly for 12 years.
n = 4 * 12 or 48 i = 4.1/4 = 1.025 = .01025 R = 15,000 n = 48 15000( (1 + 0.01025)^48 -1 / 0.01025 ) = 924136.22256731 48 * 15000 = 720,000 924136.22-720000= 204136.22 The future value of the ordinary annuity is $924136.22 The amount from contributions is $720000 and the amount from interest is $204136.22
Find the future value of the following annuity due. Then determine how much of this value is from contributions and how much is from interest. $300 deposited at the beginning of each semiannual period for 7 years at 7.54% compounded semiannually
n = 7 * 2 = 14 i = 7.54/2 = 3.77 = 0.0377 R=300 S = R [ (1+i)^n+1 - 1 / i ] - R S = 300 [ (1+0.0377)^14+1 -1 / 0.0377 ] - 300 S = 5605.43815681 Portion from Contributions = R * n = 300 * 14 = 4200 Portion from Interest= S − Portionfrom Contributions = 5605.44-4200 = 1405.44
Suppose that a family has 8 children. Also, suppose that the probability of having a girl is 1/2. Find the probability that the family has at least 7 girls.
n = 8 P = 1/2 P = (at least 7) = P( 7 successes) + P ( 8 successes) P(7 successes) use n = 8, x = 7 and p = 1/2 P (7successes 8 trials) = C(8,7) * (1/2)^7 * (1/2)^8-7 = 1/32 P (8 successes 8 trials) = C(8,8) * (1/2)^8 * (1/2)^8-8 = 1/256 1/32 + 1/256 = 9/256 The probability that the family has at least 7 girls is 9/256
A 48-year-old man puts $3000 in a retirement account at the end of each quarter until he reaches the age of 68, then makes no further deposits. If the account pays 4% interest compounded quarterly, how much will be in the account when the man retires at age 73?
n = 80 S = 3000 [ (1+0.01)^80 -1/0.01 ] S= 365014.56515827 A = 365014.57(1 + 0.01) ^ 20 A= 445387.14274989
Two dice are rolled. Find the odds of rolling the following. (a) A sum of 2 (b) A sum of 9 or 12 (c) A sum less than 9 (d) Not a sum of
(a) In reduced form, the odds in favor of rolling a sum of 2 are 1 TO 35 (b) In reduced form, the odds in favor of rolling a sum of 9 or 12 are 5 TO 31 (c) In reduced form, the odds in favor of rolling a sum less than 9 are 13 to 5 (d) In reduced form, the odds in favor of not rolling a sum of 5 ARE 8 TO 1
The numbers 1, 2, 3, 4, and 5 are written on slips of paper, and 2 slips are drawn at random one at a time without replacement. (a) Find the probability that the first number is 2, given that the sum is 3. (b) Find the probability that the first number is 4, given that the sum is 6.
(a) The probability that the first number is 2 ,given that the sum is 3 is 1/2 (b) The probability that the first number is 4 ,given that the sum is 6 is 1/4
A pair of dice is rolled. Find the probability of rolling (a) a sum not more than 3, (b) a sum not less than 8, (c) a sum between 5 and 10 (exclusive).
(a) How many possible outcomes are there from rolling two dice? 36 (A) 3 = > 3/36 = > 1/12 (b) 15 = 15/36 = 5/12 (C) 20 = 20/36 => 5/9
Suppose three balls are drawn (with replacement) from a bag that contains 3 white balls and 2 black balls. The number of black balls is counted. Complete parts (a) and (b) below.
(b) What is the expected number of black balls in the sample? 1.20
Find the expected value of the random variable.
E(x) = 2(0.2)+3(0.4)+4(0.3)+5(0.1) = 3.3
A biologist is attempting to classify 47,000 species of insects by assigning 3 initials to each species. Is it possible to classify all the species in this way? If not, how many initials should be used?
Thus, in order to classify 47,000 species, the biologist should use at least 4 initials.
The numbers 1, 2, 3, 4, and 5 are written on slips of paper, and 2 slips are drawn at random one at a time without replacement. (a) Find the probability that the first number is 4, given that the sum is 9. (b) Find the probability that the first number is 3, given that the sum is 4
(a) 1/2 (b) 1/2
The accompanying table gives a recent estimate (in thousands) of the smoking status among persons 25 years of age and over and their highest level of education in a certain state. Complete parts a through e.
P(current smoker)=. 2090 P(less than a high school diploma)=. 1425 P(current smoker and less than a high school diploma)=. 0406 P(current smoker given less than a high school diploma)=. 2851 Are the events "current smoker" and "less than a high school diploma" independent events? No
In February, a major airline had 77.3% of their flights arrive on time. Assume that the event that a given flight arrives on time is independent of the event that another flight arrives on time. a. A writer plans to take three separate flights for her publisher next month. Assuming the airline has the same on-time performance as in February, what is the probability that all three flights arrive on time? b. Discuss how realistic it is to assume that the on-time arrivals of the different flights are independent.
a. (Simplify your answer. Round to four decimal places as needed.) 77.3/100 = 0.773 => 0.773*0.773*0.773= 0.461889917 . The probability is 0.4619 b. How realistic is it to assume that the on-time arrivals of the different flights are independent? It is not very realistic since the time of arrival of one flight might affect the time of arrival of another flight for the same airline.
On a ballot, the candidates for each office are listed together under the office for which they are running. In an election with 6 candidates for one office and 4 candidates for another office, how many different versions of the ballot are possible?
6! = 720 4!= 24 720 * 24 * 2 = 34560 The number different ballot versions is 34560
Find the present value of an ordinary annuity with deposits of $16,462 quarterly for 5 years at 6.0% compounded quarterly. What is the present value?
6% / 4 = 1.5 / 100 = 0.015 so i = 0.015 4 * 5 = 20 so n = 20 R = 16462 16462 [ 1 - ( 1 + 0.015 )^ -20 / 0.015 ] 282630.13168001
Determine whether the following pair of events are dependent or independent. Event E is the event that a resident of Alabama lives in Birmingham, and F is the event that a resident of Alabama lives in either Birmingham or Huntsville.
Are the events dependent or independent? Therefore, the two events are dependent. Dependent.
Evaluate the combination. C(12,5)
C(12,5) = 792 How to do it CHEGG link: https://www.chegg.com/homework-help/evaluate-combinationc-12-5-chapter-8.2-problem-4e-solution-9780321748676-exc
Evaluate C(b,(b−1)).
C(b,(b−1))=b
Evaluate the combination. C(z,z)
C(z,z) = 1
Decide whether the exercise involves permutations or combinations, and then solve the problem. In a club with 8 male and 12 female members, how many 5-member committees can be chosen that have (a) all men? (b) all women? (c) 3 men and 2 women?
Combinations problem 8 nCr 5 = 56 ALL MALE COMMITTEES 12 nCr 5 = 792 ALL FEMALE COMMITTEES There can be 3696 5-member committees with 3 men and 2women.
A factory tests a random sample of 18 transistors for defects. The probability that a particular transistor will be defective has been established by past experience as 0.05. What is the probability that there are no defective transistors in the sample?
The probability that there are no defective transistors in the sample is about 0.3972 https://www.chegg.com/homework-help/quality-control-factory-tests-random-sample-20-transistors-d-chapter-8.4-problem-37e-solution-9780321748676-exc
The probability that a birth will result in twins is 0.015. Assuming independence (perhaps not a valid assumption), what is the probability that out of 150 births in a hospital, there will be exactly 4 sets of twins?
The probability that there will be exactly 4 setsof twins is 0.1129 150!/(150-4)!4! * 0.015^4 * (1-0.015) ^ 150 - 4 150!/(150-4)!4! = 20260275 20260275 * 0.015^4 * (1-0.015) ^ 150 - 4 = 0.1129
Mixed in a drawer are 2 blue socks, 6 white socks, and 6 gray socks. You pull out two socks, one at a time, without looking. Find the probability of getting 2 socks of the same color
Total socks available = 2 + 6 + 6 = 14 Hence, P(Getting 2 socks of same color) = P(2 Blue) + P(2 White) + P(2 Grey) = (2/14)*(1/13) + (6/14)*(5/13) + (6/14)*(5/13) = 31/91 = 0.34066
Mixed in a drawer are 8 blue socks, 6 white socks, and 2 gray socks. You pull out two socks, one at a time, without looking. Find the probability of getting 2 socks of the same color.
Total socks available = 8 + 6 + 2 = 16 Hence, P(Getting 2 socks of same color) = P(2 Blue) + P(2 White) + P(2 Grey) = (8/16)*(7/15) + (6/16)*(5/15) + (2/16)*(1/15) = 11/30 = 0.36
The number of ways to pick 6 different numbers from 1 to 46 in a state lottery is 9,366,819. Assuming order is unimportant, what is the probability of picking exactly 5 of the 6 numbers correctly?
USE nCr calculator 6 nCr 5 = 6 40 nCr 1 = 40 (6,5) (40,1) / 9366819 = x 240 / 9366819 = 2.562 × 10^ −5 The probability of selecting exactly 5 of the 6 numbers drawn is 2.562 × 10^ −5
An elevator has 4 passengers and stops at 6 floors. It is equally likely that a person will get off at any one of the 6 floors. Find the probability that at least 2 passengers leave at the same floor.
USE nPr calculator 6 nPr 4 =360 6^4 = 1296 360/1296 = .27777777 1-.27777 = 0.7222 The probability is 0.7222
The table shows frequencies for red-green color blindness, where M represents "person is male" and C represents "person is color-blind." Use this table to find P(M'|C).
P(M'|C)=. 163 M' [intersection M' & C ] - C (Totals C .007(M')/.043(C Totals) = .1627906977
If A and B are events such that P(A)=0.6 and P(A∪B)=0.7, find P(B) when (a) A and B are mutually exclusive; (b) A and B are independent.
0.7=0.6+P(B)−P(A∩B) 0.7=0.6+P(B) 0.7-0.6 0.1=P(B) (a) If A and B are mutually exclusive, then P(B)=0.1 0.7=0.6+P(B)-0.6*P(B) (b) If A and B are independent events, then P(B)=0.25
Two cards are drawn without replacement from an ordinary deck. Find the probability that two diamonds are drawn.
1/17
Suppose that a family has 4 children. Also, suppose that the probability of having a girl is 12. Find the probability that the family has the following children. Exactly 3 girls and 1
1/4 4 (1/2)^4 = 0.25 = 1/4
For many years, a state used 1 letter followed by 4 digits on its automobile license plates. Complete parts (a) through (c) below.
(a) How many different license plates are possible with this arrangement? 2.600 x 10^5 (b) When the state ran out of new numbers, the order was reversed to 4 digits followed by 1 letter. How many new license plate numbers were then possible? 2.600 x 10^5 (c) Eventually, the numbers described in part (b) were also used up. The state then issued plates with 1 letter followed by 1letterand then 4 digits. How many new license plate numbers will this provide? 6.760 x 10^6
Two cards are drawn without replacement from an ordinary deck. Find the probability that two black cards are drawn.
25/102
A certain country's postal service currently uses 4-digit zip codes in most areas. How many zip codes are possible if there are no restrictions on the digits used? How many would be possible if the first number could not be 0?
How many zip codes are possible if there are no restrictions on the digits used? 10x10x10x10 = 10^4 = 10000 How many would be possible if the first number could not be 0? 9x10x10x10 = 9000
Evaluate the combination. C(10,5)
If C(n,r) denotes the number of combinations of n elements taken r at a time, where r≤n,then C(n,r)=StartFraction n exclamation mark Over left parenthesis n minus r right parenthesis exclamation mark r exclamation mark EndFraction . Identify the values of n and r to be used in the formula. n = 10 r = 5 Thus, C(10,5)= 252.
Suppose that 6 cards are drawn from a well-shuffled deck of 52 cards. What is the probability that all 6 are black?
The CORRECT ANSWER is 253 divided by 22372 which gives you 0.0113087788 (rounded to the nearest thousandth as needed.) = 0.011
A die is rolled 11 times. Find the probability of rolling no more than 2 sixes.
The probability of rolling no more than 2 sixes is 0.7268 https://www.chegg.com/homework-help/questions-and-answers/die-rolled-11-times-find-probability-rolling-2-ones-q12112509
How many 2 card hands are possible with a 36-card deck?
The number of 2 card hands is 630. 36*35/2=630
A game consists of asking 21 questions to determine a person, place, or thing that the other person is thinking of. The first question, which is always "Is it an animal, vegetable, or mineral?", has three possible answers. All other questions must be answered "Yes" or "No." How many possible objects can be distinguished in this game, assuming that all 21 questions are asked? Are 21 questions enough?
The number of possible objects that can be distinguished is 3145728.Since the number of possible objects a person can think of is greater than this amount, 21 questions is not enough to distinguish all possible objects. There is : 3 { 21 } so 3 x 2 ^ 20 = 3145728
From a group of 7 newly hired office assistants, 4 are selected. Each of these 4 assistants will be assigned to a different manager. In how many ways can they be selected and assigned?
The office assistants can be selected and assigned in 840 different ways. 7! / 4!3! x 4! = 840 { got the 3 by subtracting 7 - 4 = 3 }
Both of a certain pea plant's parents had a gene for red and a gene for white flowers. If the offspring has red flowers, find the probability that it combined a gene for a red and a gene for white (rather than 2 for red.) Note that the gene for red is dominant over the gene for white.
The probability 2/3
If the letters c, o, m, m, i, t, t, e, and e are chosen at random, what is the probability that they spell the word "committee"?
x
In a club with 13 male and 7 female members, a 3-member committee will be randomly chosen. Find the probability that the committee contains all men
The probability that it will consist of all men is 0.2509
Suppose that a family has 7 children. Also, suppose that the probability of having a girl is 12. Find the probability that the family has at least 6 girls.
The probability that the family has at least 6 girls is 1/16
Complete parts a, b, and c for the loan described below. (a) Find the payment necessary to amortize the loan. (b) Find the total payments and the total amount of interest paid based on the calculated annual payments. (c) Create an amortization table and find the total payments and total amount of interest paid based upon the amortization table. A 4% loan of $40,000 compounded annually, with 5 annual payments.
(a) R = Pi/ 1 - (1+i)^-n i = 0.04 40000(0.04)/1 - (1.04)^-5 = 8985.08453972 (a) 8985.08 Total amount of payments =R•n =8985.08*5 =44925.4 (b) 44925.4 Total interest paid =R•n-P =44925.4-40000 =4925.4 (c) 4925.4 Amortization tablee EXCELL (c) Based on an amortization table, the total amount of the payments is $44925.4 Based on an amortization table, the total amount of interest paid is $4925.42
In the famous Traveling Salesman Problem, a salesman starts in any one of a set of cities, visits every city in the set once, and returns to the starting city. He would like to complete this circuit with the shortest possible distance. Complete parts (a) through (c) below.
(a) Suppose the salesman has 14 cities to visit. Given that it does not matter what city he starts in, how many different circuits can he take? 14! = 8717829120 He can take 8.718 × 10^10 different circuits. (b) The salesman decides to check all the different paths in part (a) to see which is shortest, but realizes that a circuit has the same distance whichever direction it is traveled. How many different circuits must he check? 14!/2! = 43589145600 = 4.359 x 10^10 (c) Suppose the salesman has 80 cities to visit. Would it be feasible to have a computer check all the different circuits? Explain your reasoning. No, it would not be feasible. The total number of circuits is so large, that it would take an unreasonable amount of time to check all the different circuits.
An endurance contest is being held with two independent groups of 8 participants. Individual participants in the contest drop out before the end of the contest with probability 0.1 (independently of other participants). What is the probability that at least 7 participants complete the endurance contest in one of the two groups, but not in both groups?
0.3039 how to do it: To determine the probability that at least 19 participants complete the endurance contest in one of the two groups, but not in both groups, first find the probability that at least 19 participants complete the contest in one of the groups. Notice that in the problem statement the probability of an individual not completing the contest is given as 0.08. Therefore, the probability of an individual completing the contest is 0.92. To find the probability that at least 19 participants complete the contest in one of the groups, find the probability that 19 participants complete the contest and the probability that all 20 participants complete the contest and add them together. The probability that 19 participants complete the contest is C(20,19)•(.92)19•(.08)1≈0.3282. The probability that all 20 participants complete the contest is C(20,20)•(.92)20•(.08)0≈0.1887. Therefore the probability that at least 19 participants complete the contest in one of the groups is approximately 0.3282+0.1887, or 0.5169. Now calculate the probability for the other group. The problem statement says that one group has at least 19 participants who complete the contest, but not both. This means that the other group must have less than 19 participants who complete the contest. In this problem the event "less than 19 participants complete the contest" is the complement of the event "at least 19 participants complete the contest." Recall that if E and E′ are complements of each other, then P(E′)=1−P(E). Now, use the formula for the probability of a complement to find the probability that less than 19 participants complete the contest. P(less than 19)≈1−0.5169, or 0.4831 Therefore, the probability for the second group is approximately 0.4831. Use the multiplication principle to find the probability of less than 19 participants completing the contest in one of the groups and at least 19 participants completing the contest in the other group. 0.5169•0.4831≈0.2497 Since either group could have been the group to have less than 19 participants complete the contest, there are, in fact, two ways the event could occur. Therefore, the probability that at least 19 participants complete the endurance contest in one of the two groups, but not in both groups is approximately 2•0.2497=0.4994.
Two cards are drawn without replacement from an ordinary deck. Find the probability that two face cards are drawn.
11/221
The probability that a male will be color-blind is 0.034. Find the probabilities that in a group of 49 men, the following will be true. (a) Exactly 5 are color-blind. (b) No more than 5 are color-blind. (c) At least 1 is color-blind.
1906884*(0.034)^(5)*(0.966)^(49-5) = 0.0189 (a) P(exactly 5)=0.0189 (b) P(no more than 5)=0.9938 (c) P(at least 1)=0.8164 https://www.chegg.com/homework-help/questions-and-answers/probability-male-color-blind-048-find-probabilities-group-55-men-following-true-exactly-5--q9658651
Suppose that 3 cards are drawn from a well-shuffled deck of 52 cards. What is the probability that all 3 are black?
2/17 https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.628643.html
Suppose that 6 cards are drawn from a well-shuffled deck of 52 cards. What is the probability that all 6 are red?
253/22372 https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.990554.html https://www.chegg.com/homework-help/questions-and-answers/suppose-6-cards-drawn-well-shuffled-deck-52-cards-probability-6-red-4-probability-round-si-q36863033
A factory tests a random sample of 27 transistors for defects. The probability that a particular transistor will be defective has been established by past experience as 0.05. What is the probability that there are no defective transistors in the sample?
27 nCr 27 = 1 The probability that there are no defective transistors in the sample is about .2503 1*.95^27 * (1-.95)^27-27 = 0.2503440897
Find the monthly house payments necessary to amortize a 9.6% loan of $269,000 over 15 years.
2825.22
A doctor is studying the relationship between blood pressure and heartbeat abnormalities in her patients. She tests a random sample of her patients and notes their blood pressures (high, low, or normal) and their heartbeats (regular or irregular). She finds that: (i) 13% have high blood pressure. (ii) 39% have low blood pressure. (iii) 12% have an irregular heartbeat. (iv) Of those with an irregular heartbeat, one-third have high blood pressure. (v) Of those with normal blood pressure, one-eighth have an irregular heartbeat. What portion of the patients selected have a regular heartbeat and low blood pressure?
37%
The coach of a softball team has 8 good hitters and 7 poor hitters. He chooses 4 hitters at random. Complete parts (a) through (c) below.
392 70 462
A marble is selected at random from a jar containing 4 red marbles, 2 yellow marbles, and 5 green marbles.
4/11
Evaluate the combination. C(40,28)
5.587 x 10 ^ 9 How to do it chegg link: https://www.chegg.com/homework-help/evaluate-combinationc-44-20-chapter-8.2-problem-5e-solution-9780321748676-exc
Suppose that a family has 4 children. Also, suppose that the probability of having a girl is 12. Find the probability that the family has at least 3 girls.
5/16
An insurance company has written 50 policies for $100,000, 100 policies for $30,000, and 250 policies for $15,000 for people of age 20. If experience shows that the probability that a person will die at age 20 is 0.0012, how much can the company expect to pay out during the year the policies were written?
50(100000)(0.0012)+100(30000)(0.0012)+250(15000)(0.0012) = 14100 The company can expect to pay out $14100 over the year after the policies were written. https://www.chegg.com/homework-help/questions-and-answers/insurance-company-written-50-policies-125-000-100-policies-55-000-250-policies-5000-people-q5580176
Mixed in a drawer are 6 blue socks, 4 white socks, and 6 gray socks. You pull out two socks, one at a time, without looking. Find the probability of getting 2 socks of the same color.
6 nCr 2 = 15 4 nCr 2 = 6 6 nCr 2 = 15 all equals 36 16 nCr 2 = 120 36/120 = 3/10 https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.446250.html
Decide whether the exercise involves permutations or combinations, and then solve the problem. In a club with 8 male and 12 female members, how many 7-member committees can be chosen that have (a) all men? (b) all women? (c) 4 men and 3 women?
Does the problem involve permutations or combinations? Combinations (a) There can be 8 all male committees. (b) There can be 792 all female committees. (c) There can be 15400 7-member committees with 4 men and 3 women.
Decide whether the exercise involves permutations or combinations, and then solve the problem. Marbles are being drawn without replacement from a bag containing 11 marbles. (a) How many samples of 4 marbles can be drawn? (b) How many samples of 3 marbles can be drawn? (c) If the bag contains 4 yellow, 2 white, and 5 blue marbles, how many samples of 4 marbles can be drawn in which all marbles are blue?
Does the problem involve permutations or combinations? Combinations (a) How many samples of 4 marbles can be drawn? 330 samples (b) How many samples of 3 marbles can be drawn? 165 samples (c) If the bag contains 4 yellow, 2 white, and 5 blue marbles, how many samples of 4 marbles can be drawn in which all marbles are blue? 5 samples
Find the expected value for the random variable x having the probability function shown in the graph.
E(x) = 30 10(0.05)+20(0.1)+30(0.7)+40(0.1)+50(0.05)
The given table indicates the transactions at a bank for one teller for one day. Letting C represent "cashing a check" and D represent "making a deposit," express PC′|D in words and find its value. Transaction Cash Check No Check Totals Make Deposit 60 30 90 No Deposit 24 14 38 Totals 84 44 128
Express PC′|D in words. The probability of a customer not cashing a check, given that the customer makes a deposit Find PC′|D. 30 / 90 = 0.33
In a game of musical chairs, 5 children will sit in 4 chairs arranged in a row (one will be left out). In how many ways can this happen, if we count rearrangements of the children in the chairs as different outcomes?
Is this problem a permutation or a combination? Permutation There are 120 different outcomes. { 5 x 4! = 120 }
Each day Donna and Mary toss a coin to see who buys the other person coffee ($1.47 a cup). One tosses and the other calls the outcome. If the person who calls the outcome is correct, the other buys the coffee; otherwise the caller pays. Assume that an honest coin is used, that Mary tosses the coin, and that Donna calls the outcome. Find Mary's expected payback. Is this a fair game?
Mary's expected payback is $0 Therefore, Mary's expected payback is $0. A fair game is a game with an expected value of 0. Because Mary's expected payback of $0, this is a fair game. Is this a fair game? YES
A bicycle factory runs two assembly lines, A and B. 92% of line A's products pass inspection and 88% of line B's products pass inspection. 60% of the factory's bikes come off assembly line B and the rest come off line A. Find the probability that one of the factory's bikes passed inspection and came off assembly line A. A B passes 92 88 produce 40 60
Multiply the probability that the bike came off assembly line A by the probability that a bike from assembly line A passed inspection. .92*.40= .368 The probability is .368
If the letters g, i, g, g, l, i, n, and g are chosen at random, what is the probability that they spell the word "giggling"?
Note: There is an extra i in the letters g,i,g,g,i,l,i,n,g which is highlighted in bold. The answer discards it. The word giggling has 4 g's, 2 i's and 1 each of l and n - 8 in all. Number of different words that can be formed = 8! / (4! * 2! * 1! * 1!) = 840. Of these giggling is one word. => Probability = 1/840 = 0.0012.
For many years, a state used 4 letters followed by 2 digits on its automobile license plates. Complete parts (a) through (c) below.
Number of letter options = 26 Number of digit options = 10 a) Number of license plates possible with 4 letters followed by 2 digits = 26^4 x 10^2 = 4.570 x 10^7 b) Number of new number plates (2 digits followed by 4 letters) = 10^2 x 26^4 = 4.570 x 10^7 c) Number of new plates with 1 letter followed by 4 letters and then 2 digits = 26 x 26^4 x 10^2 = 1.188 x 10^9
A die is rolled 12 times. Find the probability of rolling exactly 1 four.
Number of outcomes on roll of a die =6 P(getting a four on roll of a die), p = 1/6 q = 1 - p = 5/6 Number of trials, n = 12 Binomial distribution: P(X) = nCx px qn-x P(rolling exactly 1 four), P(X = 1) = 12C1 x (1/6)1 x (5/6)11 = 0.2692 https://www.chegg.com/homework-help/questions-and-answers/die-rolled-12-times-find-probability-rolling-exactly-1-four-q32753444
Make a probability distribution for the given set of events. The sums that appear when two fair four-sided dice (tetrahedrons) with sides 1, 2, 3, and 4 are tossed Complete the probability distribution below.
P (A) = number of ways A can occur / total number of possible outcomes 2 = 1/16 3 = 1/8 4 = 3/16 5 = 1/4 6 = 3/16 7 = 1/8 8 = 1/16
The table shows relative frequencies for red-green color blindness in one doctor's practice. M represents 'person is male' and C represents 'person is color-blind'. Use this table to find the probability P(C).
P(C)= 0.0552 P[totals] row(c)
Suppose that 10% of a certain batch of calculators have a defective case, and that 17% have defective batteries. Also, 3% have both a defective case and defective batteries. A calculator is selected from the batch at random. Find the probability that the calculator has a good case and good batteries
P(E) =0.1 P(F)= 0.17 P(E N F ) =0.03 = 0.1 + 0.17 - 0.03 = 0.24 1-0.24 = 0.76
A bridge hand is made up of 13 cards from a deck of 52. Find the probability that a hand chosen at random contains at least 1 seven.
Prob for a king = 4/52 = Prob that 13 cards contain atleast 1 king = 1-Prob that 13 cards do not have a king = 1- = 1- = 0.6962 The probability that a bridge hand chosen at random contains at least 1 seven is 0.6962 https://www.chegg.com/homework-help/questions-and-answers/bridge-hand-made-13-cards-deck-52-find-probability-hand-chosen-random-contains-least-1-kin-q7710283
A 43-year-old man puts $1500 in a retirement account at the end of each quarter until he reaches the age of 64, then makes no further deposits. If the account pays 4% interest compounded quarterly, how much will be in the account when the man retires at age 69?
S = R [ (1+i)^n - 1 / i ] n = 64 - 43 = 21 21 * 4 = 84 S = 1500 [ (1+0.01)^84 -1/0.01 ] S = 196008.41160605 A = P(1+i)^n 20 = (69-64)*4 A = 196008.41( 1 + 0.01)^20 A = 239167.51
An insurance company has written 100 policies for $75,000, 500 policies for $55,000, and 1000 policies for $7500 for people of age 20. If experience shows that the probability that a person will die at age 20 is 0.0015, how much can the company expect to pay out during the year the policies were written?
The company can expect to pay out $6375063750 over the year after the policies were written. https://www.chegg.com/homework-help/questions-and-answers/insurance-company-written-100-policies-75-000-500-policies-25-000-1000-policies-10-000-peo-q15697365
A committee of 3 members is selected from a club made up of 6 junior members and 24 senior members. What is the expected number of juniors in the committee?
The expected number of juniors in the committee is 0.60 https://www.chegg.com/homework-help/questions-and-answers/committee-3-members-selected-club-made-6-junior-members-24-senior-members-expected-number--q36539041
A contest at a fast-food restaurant offered the cash prizes and probabilities of winning on one visit shown to the right. Suppose you spend $1 to buy a bus pass that lets you go to 28 different restaurants in the chain and pick up entry forms. Find your expected value.
The expected value is 0.22
Find the compound amount for the deposit and the amount of interest earned. $270 at 6.4% compounded semiannually for 14 years
The principal is $270, so P=270. The interest rate needs to be expressed as a decimal, so r=0.064. The amount is compounded semiannually, so m=2. The amount is compounded for 14 years, so t=14. the value of i is 0.064/2 and the value of n is 2*14 or 28 A = P(1 + i)^n A= 270(1+0.064/2)^28 A= 652.2217664 (round to the nearest cent as needed) 652.22 The compound amount after 14years is $652.22 To find the amount of interest earned, subtract the initial deposit from the compound amount. 652.2217664 - 270 = 382.2217664 (round to the nearest cent as needed) 382.22 The amount of interest earned is $382.22
If two fair dice are rolled, find the probability of the following result. A double, given that the sum was 7.
The probability is 0.
90% of Dr.William's patients end up with20-30 vision or better. Find the probability that exactly 5 of her next 8 patients end up with20-30 vision or better.
The probability is 0.0331 this how to do it https://www.chegg.com/homework-help/questions-and-answers/90-dr-william-s-patients-end-20-30-vision-better-find-probability-exactly-5-next-8-patient-q33657573
A survey finds that customers are charged incorrectly for 2 out of every 30 items, on average. Suppose a customer purchases 15 items. Find the following probability. A customer is not charged incorrectly for any item.
The probability is 0.3553 this is how to do it https://www.chegg.com/homework-help/questions-and-answers/survey-finds-customers-charged-incorrectly-1-every-30-items-average-suppose-customer-purch-q34928590
In a particular card game, each player begins with a hand of 3 cards, and then draws 5 more. Calculate the probability that the hand will contain one pair (2 cards of one value, with the other cards of 6 different values).
The probability is 0.3923
In a particular card game, each player begins with a hand of 2 cards, and then draws 5 more. Calculate the probability that the hand will contain one pair (2 cards of one value, with the other cards of 5 different values).
The probability is 0.4728 Let S be a sample space of equally likely outcomes, and let event E be a subset of S. Then the probability that event E occurs is given by the formula P(E)=n(E)n(S). In this case, n(E) is the number of ways a hand can be drawn containing exactly one pair, and n(S) is the number of possible hands. To find n(E), first use the multiplication principle and combinations to find the number of ways to draw the pair. Then find the number of ways of drawing the other cards and multiply. Use the combinations formula to find n(S).
Suppose a box contains 3 red and 3 blue balls. A ball is selected at random and removed, without observing its color. The box now contains either 3 red and 2 blue balls or 2 red and 3 blue balls. Complete parts (a) through (c) below.
The probability that a red ball was initially removed is 0.9193 The probability that a red ball was initially removed is 0.9924 This is wrong because Ray performed the experiment more times, which makes it more accurate.
A recent study found that 81% of breast-cancer cases are detectable by mammogram. Suppose a random sample of 13 women with breast cancer are given mammograms. Find the probability that all of the cases are detectable, assuming that detection in the cases is independent.
The probability that all of the cases of breast cancer are detectable is about 0.0646 C(13,13)*0.81^13*(1-0.81)^13-13 = 0.0646
In a particular card game, each player begins with a hand of 3 cards, and then draws 4 more. Calculate the probability that the hand will contain one pair (2 cards of one value, with the other cards of 5 different values).
The probability is 0.4728 First find n(S), the number of 7-card hands. Choose the correct expression below. C(52,7) due to 52 cards in deck and 3+4 = 7 C(52,7) = 133784560 n (S) = 133784560 To find n(E), first determine how many ways there are for two cards to make up the hand's single pair. Using the multiplication principle, write the product of the number of ways to choose a value and, once a value is chosen, the number of ways to choose 2 cards with that value. Choose the correct expression below. n(one pair)= 13 * C(4,2) 13 * 6 = 78 Now determine how many possibilities there are for the 5 cards making up the balance of the hand. Note that there are only 12 remaining values to choose from and that each of the 5 values could be from any of the 4 suits. Choose the correct expression below. n(other 5)=C(12,5)•4^5 792 * 1024 = 811008 Therefore, there are n(E)=13•C(4,2)•C(12,5)•4^5 hands of 7ncards containing exactly one pair. n (E) = 78 * 811008 = 63258624 P (E) = n (E) / n (S) = X P ( E) = 63258624 / 133784560 = .4728394966 (Round to four decimal places as needed.) The probability is 0.4728
In a particular card game, each player begins with a hand of 2 cards, and then draws 4 more. Calculate the probability that the hand will contain one pair (2 cards of one value, with the other cards of 4 different values).
The probability is 0.4855
An elevator has 6 passengers and stops at 14 floors. It is equally likely that a person will get off at any one of the 14 floors. Find the probability that at least 2 passengers leave at the same floor.
The probability is 0.7128
A bag contains 6 red balls and 4 blue balls. If 5 balls are selected at random, find the probability of selecting 5 red balls.
The probability is 1/42.
Suppose that 3 cards are drawn from a well-shuffled deck of 52 cards. What is the probability that all 3 are diamonds?
The probability is 11/850
A bag contains 5 red apples and 6 yellow apples. A sample of 3 apples is drawn. Find the probability that the sample contains more red than yellow apples.
The probability is 14/33
Twenty-six slips of paper are each marked with a different letter of the alphabet and placed in a basket. A slip is pulledout, its letter recorded(in the order in which the slip wasdrawn), and the slip is replaced. This is done 4 times. Find the probability that the word Ball is formed. Assume that each letter in the word is also in the basket.
The probability is P(E)= 2.188 × 10^−6 26 raises ^ 4 = 456976 1/456976 = 2.188298729e-6
If you are dealt 4 cards from a shuffled deck of 52 cards, find the probability that all 4 cards are queens or jacks.
The probability is P(E)= 2/7735
Suppose that 5 cards are drawn from a well-shuffled deck of 52 cards. What is the probability that all 5 are not red?
The probability is P(E)= 253/9996
If you are dealt 4 cards from a shuffled deck of 52 cards, find the probability that all 4 cards are kings.
The probability is 1/270725 # of ways to succeed: 1# of possible outcomes: 52C4 = 270,725--------------P(4 queens) = 1/270725 = 0.00000369... Cheers, Stan H.============ how to do it: https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.651431.html
A bag contains 7 red apples and 4 yellow apples. A sample of 3 apples is drawn. Find the probability that the sample contains more red than yellow apples.
The probability is 119/165 This is how to do it CHEGG: https://www.chegg.com/homework-help/questions-and-answers/basket-contains-7-red-apples-4-yellow-apples-sample-3-apples-drawn-find-probability-red-ye-q1059493
If two fair dice are rolled, find the probability that the sum of the dice is 5, given that the sum is greater than 4.
The probability is 2/15
A shipment of 10 printers contains 2 that are defective. Find the probability that a sample of size 2, drawn from the 10, will not contain a defective printer.
The probability is 28/45 how to do it: https://www.chegg.com/homework-help/questions-and-answers/shipment-10-printers-contains-2-defective-find-probability-sample-size-2-drawn-10-contain--q20129273
Two crows, 7 bluejays, and 8 starlings sit in a random order on a section of telephone wire. Find the probability that birds of a feather flocktogether, thatis, that all birds of the same type are sitting together.
The probability is 66.856 × 10 ^ −6 2! 7! 8! there is 3! different types of birds 2! 7! 8! = 17! so 3!2!7!8! / 17! = 66.856 × 10 ^ −6 The probability is 66.856 × 10 ^ −6 . It's simple all you have to do is Two crows , 7 blue jays and 8 starlings. Basically 2! 7! and 8! and there is 3 different types of bird so 3! . Add 2 + 7 + 8 you get total of 17! Equation = 3! 2! 7! 8! divide by 17! = 66.856 × 10 ^ −6 { BERT }
An experiment consists of dealing 5 cards from a standard 52-card deck. What is the probability of being dealt a 4, 5, 6, 7, 8, all in the same suit?
The probability of being dealt a 4, 5, 6, 7, 8, all in the same suit is 0.0000015 https://www.chegg.com/homework-help/questions-and-answers/experiment-consist-dealing-5-cards-standard-52-card-deck-probability-dealt-10-jack-queen-k-q20129457
The number of ways to pick 6 different numbers from 1 to 43 in a state lottery is 6,096,454. Assuming order is unimportant, what is the probability of picking exactly 3 of the 6 numbers correctly?
The probability of selecting exactly 3 of the 6 numbers drawn is 2.549×10^−2
A reader wrote to a column in a magazine, "A dozen glazed doughnuts are riding on the answer to this question: Are the odds of winning in a lottery drawing higher when picking 5 numbers out of 48 or when picking 4 numbers out of 51?" Calculate each probability to answer the question.
The probability of winning the lottery when picking 5 numbers out of 48 is 1/1712304 The probability of winning the lottery when picking 4 numbers out of 51 is 1/249900 The probability of winning the lottery when picking 4 numbers out of 51 is higher. https://www.chegg.com/homework-help/questions-and-answers/reader-wrote-column-magazine-dozen-glazed-doughnuts-riding-answer-question-odds-winning-lo-q44913645
The following table lists the number of passengers who were on a cruise ship that sank and the number of passengers who survived, according to class of ticket. Use this information to complete parts (a) through (g).
What is the probability that a randomly selected passenger was second class? .2729 What is the overall probability of surviving? .3802 What is the probability of a second-class passenger surviving? .3213 What is the probability of a child who was also in the second class surviving? 1 Given that the survivor was from second class, what is the probability that the survivor was a child? .2241 Given that a woman survived, what is the probability that she was in third class? .2678 Are the events third-class survival (E) and woman survival (F) independent events? What does this imply? The events are not independent events because the probability of third-class survival, given that the passenger is a woman, is not equal to the probability of third-class survival. What does this imply? The probability that a third-class passenger survived depended on whether that passenger was a man or a woman.
According to an airline, a particular flight is on time 88% of the time. Suppose 47 flights are randomly selected and the number of on time flights is recorded. Find the probabilities of the following events occurring. a. All 47 flights are on time b. Between 41 and 43 flights (inclusive) are on time
a. The probability that all 47 flights are on time is 0.0025. (Round to four decimal places as needed.) b. The probability that between 41 and 43 flights, inclusive, are on time is 0.4991 https://www.chegg.com/homework-help/questions-and-answers/11-according-airline-particular-flight-time-88-time-suppose-47-flights-randomly-selected-n-q50704900
Five cards are marked with the numbers 1, 2, 3, 4, and 5, then shuffled, and two cards are drawn. a. How many different 2-card combinations are possible? b. How many 2-card hands contain a number less than three?
a. The total number of 2-card combinations is 10. b. The number of 2-card hands that contain a number less than three is 7.
A company sells regular hamburgers as well as a larger burger. Either type can include cheese, relish, lettuce, tomato, mustard, or ketchup. a. How many different hamburgers can be ordered with exactly 4 extras? b. How many different regular hamburgers can be made that use any 4 of the extras? c. How many different regular hamburgers can be ordered with at least 2 extras?
a. There are 30 different hamburgers that can be made using exactly 4 of the extras. b. There are 15 different regular hamburgers that can be made using exactly 4 of the extras. c. There are 57 different regular hamburgers that can be ordered with at least 2 of the extras.
In a certain state, 37.1% of all community college students belong to ethnic minorities. Find the probabilities of the following results in a random sample of 12 of the community college students. a. Exactly 3 belong to an ethnic minority. b. Three or fewer belong to an ethnic minority. c. Exactly 6 do not belong to an ethnic minority. d. Eight or more do not belong to an ethnic minority.
a. P(3)=0.1731 (Round to four decimal places as needed.) b. P(three or fewer belong to an ethnic minority)=0.2921 (Round to four decimal places as needed.) c. P(exactly 6 do not belong to an ethnic minority)=0.1492 (Round to four decimal places as needed.) d. P(eight or more do not belong to an ethnic minority)=0.5219 (Round to four decimal places as needed.)
A raffle offers a first prize of $500 and 3 second prizes of $80 each. One ticket costs $3, and 400 tickets are sold. Find the expected payback for a person who buys 1 ticket. Is this a fair game?
he expected payback for a person who buys 1 ticket is −1.15. The raffle is not a fair game because the expected payback of the game is not equal to $0.
Suppose a fair die numbered 1 to 6 is rolled 4 times. Complete parts (a) and (b) below.
https://www.chegg.com/homework-help/questions-and-answers/suppose-fair-die-numbered-1-6-rolled-4-times-complete-parts-b-find-probability-distributio-q36539089
Amortization Table
https://www.excel-easy.com/examples/loan-amortization-schedule.html
Kristen's financial advisor has given her a list of 12 potential investments and has asked her to select and rank her favorite five. In how many different ways can she do this?
use calculator 12 nPr 5 = 95040
According to a survey, 25.9% of credit-card-holding families in a certain area hardly ever pay off the balance. Suppose a random sample of 29 credit-card-holding families is taken. Find the probability that at least 4 families hardly ever pay off the balance.
x
How many 7-digit telephone numbers are possible if the first digit cannot be four and (a) only odd digits may be used? (b) the number must be a multiple of 10 (that is, it must end in 0)? (c) the number must be a multiple of 1,000? (d) the first 3 digits are 356? (e) no repetitions are allowed?
(a) 78,125 telephone number(s) (b) 900,000 telephone number(s) (c) 9,000 telephone number(s) (d) 10000 telephone number(s) (e) 544320 telephone number(s)
How many 7-digit telephone numbers are possible if the first digit cannot be one and (a) only even digits may be used? (b) the number must be a multiple of 10 (that is, it must end in 0)? (c) the number must be a multiple of 1,000? (d) the first 4 digits are 9738? (e) no repetitions are allowed?
(a) 78,125 telephone number(s) (b) 900,000 telephone number(s) (c) 9,000 telephone number(s) (d) 10000 telephone number(s) (e) 544320 telephone number(s)
One of the few methods that can be used in an attempt to cut the severity of a hurricane is to seed the storm. In this process, silver iodide crystals are dropped into the storm. Unfortunately, silver iodide crystals sometimes cause the storm to increase its speed. Wind speeds may also increase or decrease even with no seeding. Use the table below to answer parts (a) and (b).
(a) Find the expected amount of damage under each option, "seed" and "do not seed." The expected amount of damage under the seed option is about $92.3 million. (Round to one decimal place as needed.) The expected amount of damage under the do not seed option is about $115.0 million. (Round to one decimal place as needed.) (b) To minimize total expected damage, what option should be chosen? The option to seed should be chosen because it is expected to result in less
A group of 10 workers decides to send a delegation of 5 to their supervisor to discuss their grievances. Complete parts (a) through (c) below.
(a) How many delegations are possible? There are 252 delegations possible. (b) If it is decided that a particular worker must be in the delegation, how many different delegations are possible? There are 126 delegations possible. (c) If there are 3 women and 7 men in the group, how many delegations would include at least 1 woman? There are 231 delegations possible. How to do it: https://www.chegg.com/homework-help/questions-and-answers/group-11-workers-decides-send-delegation-4-supervisor-discuss-grievances-complete-parts-c--q47947401
A lottery game requires that you pick 7 different numbers from 1 to 60. If you pick all 7 winning numbers, you win the jackpot. Complete parts (a) and (b) below.
(a) How many ways are there to choose 7 numbers if order is not important? Choose the correct answer below. There are 386,206,920 ways to choose the numbers. (b) How many ways are there to choose 7 numbers if order matters? There are 1,946,482,876,800 ways to choose the numbers.
Seven orchids from a collection of 16 are to be selected for a flower show. Complete parts (a) and (b) below.
(a) In how many ways can this be done? This can be done in 11440 ways. (b) How many ways can the 7 be selected if 2 special plants from the 16 must be included? This can be done in 2002 ways.
In placebo-controlled trials of a new drug that is prescribed to lower cholesterol, 7.4% of the patients who were taking the drug experienced nausea/vomiting, whereas 7.3% of the patients who were taking the placebo experienced nausea/vomiting. (a) If 60 patients who are taking the drug are selected, what is the probability that 10 or more will experience nausea/vomiting? (b) If 60 patients who are taking the placebo are selected, what is the probability that 10 or more will experience nausea/vomiting? (c) Since 7.4% is larger than 7.3%, do you believe that the new drug causes more people to experience nausea/vomiting than a placebo? Explain.
(a) The probability that 10 or more patients taking the drug will experience nausea/vomiting is 0.0122 (b) The probability that 10 or more patients taking the placebo will experience nausea/vomiting is 0.0111 (c) Since 7.4%is larger than 7.3%,do you believe that the new drug causes more people to experience nausea/vomiting than a placebo? Explain Yes, because the probability of experiencing nausea/vomiting is greater on the drug than the probability of experiencing nausea/vomiting on the placebo. This how to do it: https://www.chegg.com/homework-help/questions-and-answers/study-tracked-cohort-low-birth-weight-infants-twenty-years-results-study-indicated-71-low--q57153960?trackid=ncinkNSc
A study tracked a cohort of very-low-birth-weight infants for twenty years. The results of the study indicated that 71% of the very-low-birth-weight babies graduated from high school during this time period. The study also reported that 82% of the comparison group of normal-birth-weight babies graduated from high school during the same period. (a) If 40 very-low-birth-weight babies were tracked through high school, what is the probability that at least 30 will graduate from high school by age 20? (b) If 40 normal-birth-weight babies were tracked through high school, what is the probability that at least 30 will graduate from high school by age 20?
(a) The probability that at least 30 out of 40 very-low-birth-weight babies will graduate from high school by age 20 is 0.3590 (b) The probability that at least 30 out of 40 normal-birth-weight babies will graduate from high school by age 20 is 0.9084 This how you do it: https://www.chegg.com/homework-help/questions-and-answers/study-tracked-cohort-low-birth-weight-infants-twenty-years-results-study-indicated-71-low--q57153960?trackid=ncinkNSc
Decide whether the exercise involves permutations or combinations, and then solve the problem. There are 7 rotten apples in a crate of 24 apples. (a) How many samples of 3 apples can be drawn from the crate? (b) How many samples of 3 could be drawn in which all 3 are rotten? (c) How many samples of 3 could be drawn in which there are two good apples and one rotten one?
Combinations (a) How many samples of 3 apples can be drawn from the crate? 2024 samples (b) How many samples of 3 could be drawn in which all 3 are rotten? 35 samples (c) How many samples of 3 could be drawn in which there are two good apples and one rotten one? 952 samples
A biologist is attempting to classify 800 species of insects by assigning 2 initials to each species. Is it possible to classify all the species in this way? If not, how many initials should be used?
No, it is not possible to classify all the species by assigning 2initials to each species. The biologist should use at least 3 initials instead.
A biologist is attempting to classify 900 species of insects by assigning 2 initials to each species. Is it possible to classify all the species in this way? If not, how many initials should be used?
No, it is not possible to classify all the species by assigning 2initials to each species. The biologist should use at least 3 initials instead.
A red 8-sided die and a green 8-sided die are rolled. Let A be the event that the red die comes up even, and let B be the event that the green die comes up even. Are these events dependent or independent?
P(A)= 1/2 P(A|B)= 1/12 Are events A and B dependent or independent? Events A and B are independent because P(A|B)=P(A).
The table below lists the number of head and neck injuries for ice hockey players' exposures wearing either a full shield or half shield in a hockey league. Find the probability P(A|H)
P(A|H)= .192 A column = 28 H row total 146 28/146= .1917808219