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- Each day Mr. Samms randomly chooses 2 students from his class to serve as helpers. If there are 15 boys and 10 girls in the class, what is the probability that the two students chosen will be female?
- prob and stat cp
Whiting, indiana, leads the “Top 100 Cities with the Oldest Houses” list with the average age of houses being 66.4 years. Farther down the list resides Franklin, Pennsylvania, with an average house age of 59.4 years. Researchers selected a random sample of 20 houses in each city and obtained the following statistics. At =0.05, can it be concluded that the houses in Whiting are older?

- a box contains 25 computer diskettes, 5 of which are double density and the remiander are high density
a box contains 25 computer diskettes, 5 of which are double density and the remiander are high density. A second box contains 15 diskettes, 10 double density and 5 high density. A box is selected at random and then two diskettes are chosen randomly from the selected box. What is the probability that both diskettes are high density?

- Bruskin Associates Market Research found that 40% of Americans do not think that having a
college education is important to succeed in the business world
R. H. Bruskin Associates Market Research found that 40% of Americans do not think that having a college education is important to succeed in the business world. If a random sample of five Americans is s selected, find these probabilities. (a) Exactly two people will agree with that statement. (b) At most three people will agree with that statement. (c) At least two people will agree with that statement. (d) Fewer than three people agree with that statement.

- A college professor has been studying the relationship between the attendance habits of his students and the grades they earn in his courses
A college professor has been studying the relationship between the attendance habits of his students and the grades they earn in his courses. A breakdown of 200 of his students revealed the following: Attended regularly --- 160 Attended sometimes --- 40 Passed --- 170 and Failed --- 30 140 Passed and 20 Failed 30 Passed and 10 failed Assume that a student is chosen at random. What is the probability that the student passes the course, given that the student attended class sometimes?

- Could someone please check this and let me know if I'm doing it correctly
Could someone please check this and let me know if I'm doing it correctly? Sammy and Sally each carry a bag containing banana, chocolate bar, and a licorice stick. Simutaneously, they take out a single good item and consume it. The possible food items that Sally and Sammy consumed are as follows. Chocolate bar - chocolate bar licorice stick - chocoate bar banana - banana chocolate bar - licorice stick licorice stick - licorice stick chocolate bar - banana banana - licorice stick licorice stick - banana banana - chocolate bar Find the probabilty that at least one chocoate bar was eaten. = 5/9 Find the conditional probability. Suppose one card is selected at random from an ordinary deck of 52 playing cards. Let A = event a queen is selected. B = event a diamond is selected. Determine P (B!A) = P (13/4) correct yay for you Hi! Very nice site! Thanks you very much! RBwwgnUJe2iB Hello! Good Site! Thanks you! gekpawzwli 69% of adults favor gun licensing in general. Chose 1 adult at random. What is the probability that the selected adult does not believe in gun licensing?

- Im not sure how to start here
Im not sure how to start here. Please help. The age distribution of students at a community college is given below. Age (years) Under 21 21-25 26-30 31-35 Over 35 Number of students (f) 411 407 201 51 26 A student from the community college is selected at random. Find the conditional probability that the student is between 26 and 30 given that he or she is at least 26. Of the 278 (201+51+26) students who are 26 or older, 201 are 26-30. That is a probability of 206/278 = 74.1% I'm not sure I understand this. Why are we including 51, and 26 if there are only 201 students that are between 26-30. Where does 206 come from? Hi! Very nice site! Thanks you very much! xK6Vi84ZHm8I

- A survey of 850 students conducted in two schools showed the following results:
School A:
*Students who took the bus = 400
* Students who did not take the bus = 150
Total = 550
School B:
*Students who took the bus = 200
A survey of 850 students conducted in two schools showed the following results: School A: *Students who took the bus = 400 * Students who did not take the bus = 150 Total = 550 School B: *Students who took the bus = 200 * Students who did not take the bus = 100 Total = 300 School C: *Students who took the bus = 600 * Students who did not take the bus = 250 Total = 850 Calculate the probability that a randomly chosen student from this group of 850 students takes the bus and is from school A.

- out of 250 students interviewed at a community college, 90 were taking mathematics but not computer science, 160 were taking mathematics, and 50 were taking neither mathematics nor computer science
out of 250 students interviewed at a community college, 90 were taking mathematics but not computer science, 160 were taking mathematics, and 50 were taking neither mathematics nor computer science. Find the probability that a student chosen at random was a. taking just computer science b. taking mathematics or computer science, but not both c. taking computer science d. not taking mathematics e. taking mathematics, given that the student was taking computer science f. taking computer science, given that the student was taking mathematics g. taking mathematics, given that the student was taking computer science or mathematics h. taking computer science, given that the student was not taking mathematics i. not taking mathematics, given that the student was not taking computer science

- ) An octave contains twelve different musical notes (in Western music)
1.) An octave contains twelve different musical notes (in Western music). How many different eight note melodies can be constructed from these twelve notes if: (a) no note can be used more than once? (b) any note can be used as often as you please? ============================= Ans for a) a permutation 12P8 =19,958,400 b) a combination 12C8= 495 Is this correct? ================================== ==================================== 3. The chance that Ariel sees the movie Norbit is 45%. The chance that Brandon sees the movie is 50%. The chance that Ariel and Brandon both see the movie is 30%. (a) If Brandon sees the movie, what is the chance that Ariel also sees it? ======================================= Answer: P(A&B)=(.45)(.50)= 22.5% ======================================= ======================================= 4. In a certain election, the incumbent Republican will run against the Democratic nominee. There are three Democratic candidates, D1, D2 and D3, whose chances of gaining the Democratic nomination are .50, .35 and .15, respectively. Here are the chances that the Republican will win against each of these possible Democratic nominees: vs. D1: 0.60 vs. D2: 0.50 vs. D3: 0.40 (a) Name (but do not give) the probability formula that is needed to find the chance that the Republican will win the election. (b) Find the probability that the Republican will win the election. ===================================== not sure how to do this one. ===================================== ===================================== 5. Widgets are produced at a certain factory by each of three machines A, B and C. These machines produce 1000, 600 and 400 widgets per day, respectively. The probability that a given widget is defective is 4% for one produced by Machine A, 3% if produced by Machine B, and 2% if produced by Machine C. Suppose that the Widget Inspector selects a widget at random from the factory's widget inventory. (a) What is the probability that the widget is defective? (b) If the widget is defective, what is the chance that it came from Machine A? ====================================== Answer: a) 4+3+2=9% b) P(A&B)= 9% * (1000/2000) = 4.5% ====================================== ======================================= 6. A fair coin is flipped three times. You win $5 every time the outcome is heads. Let the random variable X represent the total number of dollars you win. (a) List the sample space. (b) Determine the probability function of X. ====================================== Answer: a) 2^3= 8 possibilities b) (.5+.5+.5)/15= 1/10 ====================================== ====================================== 7. Suppose that there are nine adjacent parking spaces in one row of a parking lot. Nine cars are to be parked by an attendant. Three are SUV’s, three are compacts, and three are expensive sports cars. Assuming that the attendant parks the cars randomly, find the chance that the three expensive sports cars are parked next to one another. ====================================== Answer: not sure how to do this one. ======================================

- You are the owner of an auto repair service
You are the owner of an auto repair service. History tells you it takes on average 45 minutes to complete a repair job. You have determined the standard deviation for a job is 6 minutes. A women comes into your shop and tells you she must leave her car for repair, and that she will be shopping at the mall across the street. She says she will be back no earlier than 38 minutes, but she absolutely must leave no later than 55 minutes to pick up her child from school. What is the probability her car will be repaired during the 38 to 55 minute timeframe?

- Reformatted so that it's easier to read
Reformatted so that it's easier to read. I have a Statistics project due tomorrow on Combinatorics. Admittedly, while my math skills aren't too bad (I get 99th percentile on standardized tests and such), it has been a while since I last took a math class after a 2-year lapse in education, so my skills are really rusty. I would greatly appreciate some help checking to see if my math is correct. I will show the work I currently have written down. My intention is not to simply ask for answers and be lazy; I am asking to see if I am on the right track. The problem is: "Suppose you and your friends are playing 5-card stud poker (standard deck of 52 playing cards). Using combinations, you want to determine what is the probability of being dealt a hand of "two-pair;" similar to the hand shown in the following figure (a hand of two-pair is two cards of one value; two cards of another value; and a single card of a third value). The paper then shows 5 specific cards: an 8 of spades, and 8 of hearts, a 3 of clubs, a 3 of diamonds, and a King of diamonds. a.) Determine the number of different ways of being dealt 5 cards in a game of poker. I wrote: nCr(52, 5) =52!/[5!(52-5)!] =52!/(5!47!) =311875200/120 =2598960 b.) Determine the number of ways in a hand of 5 cards of being dealt two cards of one value; then two cards of another value; and then a single card of a third value (see figure on last page). i) To do this, determine the number of ways of being dealt only two cards of the same value: I wrote: nCr(4, 2) =4!/[2!(4-2)!] =4!/(2!2!) [B]=6[/B] ii.) Of course this happens twice, for the other set of two cards: I think I'm sure about this one, but do I have to account for the other two cards that I drew or does that not matter? I wrote: nCr(4, 2) =4!/[2!(4-2)!] =4!/(2!2!) =6 I get confused from part iii onwards: iii.) These outcomes are for these specific card values. But remember, there are 13 different card values in a standard deck of cards. Now determine how many ways these two group values can be dealt to you: My reasoning here was that since I'm looking for an 8 of spades + 8 of hearts, and a 3 of clubs and a 3 of diamonds, and that there is 13 cards of each suit, that there are 26 possible cards for spades + hearts and 26 possible cards for clubs + diamonds, and that I'm choosing two from 26. ... so I wrote: nCr(26, 2) =26!/[2!(26-2)!] =26!/[2!24!] =325 325*2 (to account for both pairs)= 650 iv.) Finally, we determine the number of ways the single card can be dealt. Remember to keep in mind that other cards have already been dealt. My reasoning here is that since the 4 other cards have been dealt, I subtract 4 cards from 52, giving me 48. Since I am choosing 1 card (King of Diamonds) from 48, I write: nCr(48, 1) =48!/(1!(48-1)!) =48!/47! =48 v.) Finally, using the Fundamental Counting Principle we determine the different ways altogether these cards can be dealt: I have no idea what this is even asking, so I don't even know where to start. vi.) To determine the probability of this happening, divide the answer obtained from v) above by the total number of ways of being dealt a 5 card hand. I have no idea how to approach this one, since it relies on an answer from v). I hope all of you don't mind that I ask? Sorry for the long post! Don't make fun of me for failing at math.

- How do I do this probability math problem,--- In 2000, the population of a certain country was about 195 million
How do I do this probability math problem,--- In 2000, the population of a certain country was about 195 million. The overall birth rate was 18.8 births per 100. The overall death rate was 11.8 deaths per 1000. Based on births & deaths alone(not counting immigration) how much did the population of the country increase during 2000? Thank you in advance

- R
R. H. Bruskin Associates Market Research found that 40% of Americans do not think that having acollege education is important to succeed in the business world. If a random sample of five Americansis s selected, find these probabilities.(a) Exactly two people will agree with that statement.(b) At most three people will agree with that statement.(c) At least two people will agree with that statement.(d) Fewer than three people agree with that statement.

- There are 12 CD's in a bag -- 6 rap music, 3 country music, and 3 rock music
There are 12 CD's in a bag -- 6 rap music, 3 country music, and 3 rock music. Two CD's are selected from the bag. The first CD is not replaced before making the second selection. What is the total number of possible out outcomes?

- An insurer offers a health plan to the employees of a large company
An insurer offers a health plan to the employees of a large company. As part of this plan, the individual employees may choose two of the supplementary covaerges A, B and C or they may choose no supplementary coverage. The proportions of the company's employees that choose coverage A, B and C are 1/4 ,1/3 and 5/12 respectively. Determine the probability that a randomly chosen employee will choose no supplementary coverage.

- 1
1. The time it takes a student to cycle to school is normally distributed with a mean of 15 minutes and a variance of 2. The student has to be at school at 8.00a.m. At what time should the student leave her house so that she will be late only 4% of the time?2. In a study for Health Statistics, it was found that the mean height of female 20-25 years old was 64.1 inches. If height is normally distributed with a standard deviation of 2.8 inches. a)Determine the height required to be in the top 10% of all 20-25 years old females.b)A “one-size-fits all” robe is being designed that should fit 90% of 20-25 years old females; what heights constitute the middle 90% of all 20-25 years old females?3. The compressive strength of cement is assumed to be normally distributed with a mean of 8000 kilograms per sq cm and a standard deviation of 200 kilograms per sq. cm.a)Find the probability that the strength is less than 8200 kg per sq. cmb)Find the probability that the strength is between 7500 and 7700 kg per sq. cmc)Determine the value for which the probability that the strength of cement is below this value is 90%.

- probability word problems
If a jar contains seven orange,four red, and eight blue marbles. What is the probability that the marble you pick will be blue?

- what is the probability that you get a 2 and a 5 without regard to which is on which die
Roll two fair dice. what is the probability that you get a 2 and a 5 without regard to which is on which die? What is the probability of at least one 2 or one 5? What is the probability of a sum of 7? "what is the probability that you get a 2 and a 5 without regard to which is on which die? " 1/6* 1/6* 2 = 1/18 There are two possibilities both with probability 1/6 * 1/6. "What is the probability of at least one 2 or one 5?" There are two ways to do this. Let's first do it using an elementary trick: Calculate the probability that you don't satisfy this critrium and then take 1 minus this probability: NOT [at least one 2 OR one 5] You simplify this expression by using the rule that says: NOT[A OR B] = NOT[A] AND NOT[B] In other words, for [A OR B] to be not true, both A and B must be false. NOT [at least one 2 OR one 5] = Not[at least one 2] And Not[one 5] = = [No 2's] And [no 5's Or two 5's] So, you can't have any 2's and you must have either no five or two five. The "No 2's" criterium means that for each dice can only have 5 possible values. To satisfy the second constraint you must either have no 5's for which there are 4*4 = 16 possibilities (both the two and the five are illegal) or you must have two fives, this can be realized in only one way. The zero fives and the two fives are mutually exclusive so you can add up the number of ways you can realize the two criteria. So, there are 17 ways to satisfy the criterium of: [No 2's] And [no 5's Or two 5's] The probablity is thus 17/36 and the probability that this is not satisfied is thus: 1-17/36 = 19/36 A more direct way to calculate this is as follows. The probability of any given outcome is 1/36, so we can write the probability as the number of dice configurations that satisfy the criterium divided by 36. Suppose you can calculate easily thenumber of caonfigurations such that criterium A is satisfied and you can also calculate the number of configuraions such that some other criterium B is satisfied. Then, the number of ways that A or B is satisfied is given by: N[A OR B] = N[A] + N[B] - N[A AND B] N[A] is the number of configurations such that A is satisfied and N[B] is the number of configurations such that B is satisfied. Suppose that it is possible to satisfy both. Then, any configuration that satisfies both A and B will be counted both in N[A] and in N[B]. That's a double counting and to correct for that we must subtract N[A AND B] N[at least one 2 or one 5] = N[at least one 2] + N[one 5] - N[at least one 2 AND one 5] N[at least one 2] can be evaluated using this rule or using by calculating the number of configurations that don't satisfy this criterium and then subrtacting 36 from that. N[at least one 2] = N[die one shows a 2 OR die 2 shows a 2] = N[die one shows a 2] + N[die two shows a 2] - N[die one shows a 2 AND die two shows a 2] = 6+6-1 =11 Alternatively you can calculate the number of configurations in which you don't have at least one 2. Then you have 5 possiblities for both dice so there are 25 possibilites in total. And 36 - 25 = 11 So we have found that: N[at least one 2] = 11 THe next term we must evaluate is: N[one 5]. This is easy. To have one 5 one die must show a five and the other must swow something else. If the first die shows the five there are 5 possibilites for the other die. And if the second one shows the 5 there are 5 possibilites for the first die. So, there are a total of ten possibilities: N[one 5] = 10 Finally we must evaluate: N[at least one 2 AND one 5]. This is easy, because if one of the dice is 5 and you must have at least one die to show a 2, the other die must be a two. You then have two possibilities one for die one showing a two and the other one showing a five and vice versa. So, we have: N[at least one 2 AND one 5]= 2 The answer is thus: N[at least one 2 or one 5] = N[at least one 2] + N[one 5] - N[at least one 2 AND one 5] 11 + 10 - 2 = 19 and the probablity is 19/36 in agreement with what we found above.

- At a certain university, 30% of the students major in math
At a certain university, 30% of the students major in math. Of the students majoring in math, 60% are males. Of all the students at the university, 70% are males. a) What is the probability that a student selected at random in the university is male and majors in math? b) What is the probability that a randomly selected student in the university is female or majors in math or both?Based upon the information you provided, out of every 100 students you have this breakdown: male, math major: 18 female, math major: 12 male, non-math major: 52 female, non-math major: 18 therefore: a) 18% b) female: 30% math major (either [email protected]#$%^&): 30% female math major: 12% so for b), you just add them up to get 72%?