MAT217 Final Exam Word Problems Review

❗️In a certain year, 87% of all Caucasians in the U.S., 75% of all African-Americans, 75% of all Hispanics, and 76% of residents not classified into one of these groups used the Internet for e-mail. At that time, the U.S. population was 68% Caucasian, 12% African-American, and 12% Hispanic. What percentage of U.S. residents who used the Internet for e-mail were Hispanic?

11% Formula: [(75 x 12) / ((68 x 87) + (12 x 75) + 76 x8))] x 100 >>> 900/8324 = 11%

The following table shows the approximate number of males of Hispanic origin employed in the United States in a certain year, broken down by age group. Age | 15-24.9 | 25-54.9 | 55-64.9 Emp. (Thous.) 26,000 | 14,000 | 2,600 (a) Use the rounded midpoints of the given measurement classes to compute the expected value and the standard deviation of the age X of a male Hispanic worker in the United States. expected value= standard deviation= (b) In what age interval does the empirical rule predict that 68 percent of all male Hispanic workers will fall? (___, ____)

A) Expected value: 29.00 Standard Deviation: 12.12 B) (16.88, 41.12) A) Ave. age or x: | 20 | 40 | 60 Prob. of emp: | .61 | .33 | 0.6 X (prb emp): | 12.2 | 13.2 | 3.6 x̅ = sum of (x•prob) = 29 = expected value (X - x̅) | -9 | 11 | 31 (x - x̅)^2: | 81 | 121 | 961 (X)(x - x̅)^2: | 49.91 | 33.93 | 57.66 σ² = sum of (X)(x - x̅)^2 = 147 σ = 12.12 B) (( x̅ - σ) , ( x̅ - σ)) (29 - 12.12, 29-12.12)

<7.1 Samples and Spaces> A packet of gummy candy contains four strawberry gums, two lime gums, two black currant gums, and four orange gums. April May sticks her hand in and selects six at random. Complete the following sentences. (a) The sample space is the set of all sets of ______ gummy candies chosen from the packet of ______. (b) April is particularly fond of combinations of four strawberry and two black currant gums. The event that April will get the combination she desires is the set of all sets of 6 gummy candies in which four are strawberry and two are black currant.

Answer: (a) The sample space is the set of all sets of __6__ gummy candies chosen from the packet of __12__. Total # gums: 12 She selects 6 at random. (b) April is particularly fond of combinations of four strawberry and two black currant gums. The event that April will get the combination she desires is the set of all sets of 6 gummy candies in which four are strawberry and two are black currant.

According to a New York Times/CBS poll, 79% agreed that it should be the government's responsibility to provide a decent standard of living for the elderly, and 46% agreed that it would be a good idea to invest part of their Social Security taxes on their own. A) What is the smallest percentage of people who could have agreed with both statements? B) What is the largest percentage of people who could have agreed with both statements?

Answer: A) What is the smallest percentage of people who could have agreed with both statements? 25 % .46 - (1-.79) = .46 - .21= 25 B) What is the largest percentage of people who could have agreed with both statements? 46 % given

According to a New York Times/CBS poll, 53% agreed that Social Security taxes should be raised if necessary to keep the system afloat, and 42% agreed that it would be a good idea to invest part of their Social Security taxes on their own. What is the largest percentage of people who could have agreed with at least one of these statements? What is the smallest percentage of people who could have agreed with at least one of these statements?

Answer: What is the largest percentage of people who could have agreed with at least one of these statements? 95%. .53 + .42 = .95 What is the smallest percentage of people who could have agreed with at least one of these statements? 53%. Given

<7.1 Samples and Spaces> My couch potato friend enjoys sitting in front of the TV and grabbing handfuls of 5 chocolates at random from his snack jar. Unbeknownst to him, I have replaced one of the 20 chocolates in his jar with a cashew. (He hates cashews with a passion.) A) How many possible outcomes are there the first time he grabs 5 chocolates? B) How many of these include the cashew?

Answer: A) 15504 outcomes ^^^20 chocolates and 5 to choose from >>> c(20,5) = 15504 B) 3876 outcomes ^^^ 19 choices left and 4 more to choose from >>> c(19,4) = 3786

A roulette wheel has the numbers 1 through 36, 0, and 00. A bet on six numbers pays 5 to 1 (that is, if one of the six numbers you bet comes up, you get back your $1 plus another $5). How much do you expect to win with a $1 bet on six numbers? $

Answer: $-0.05 38 total numbers Pays 5 6/38 = numbers you bet coming up 32/38 = 32 numbers of the 38 besides the 6 numbers E(w) = 5 x (6/38 - 32/38) = -1/19 = -0.05

Lance the Wizard has been informed that tomorrow there will be a 60% chance of encountering the evil Myrmidons and a 20% chance of meeting up with the dreadful Balrog. Moreover, Hugo the Elf has predicted that there is a 10% chance of encountering both tomorrow. What is the probability that Lance will be lucky tomorrow and encounter neither the Myrmidons nor the Balrog?

Answer: .3 Solving for P(m ∪ B') m: P(m) = .60 B: P(B) = .20 P(m ∩ B)= .10 P(m ∪ B) = P(m) + P(B) - (m ∩ B) P(m ∪ B) = .70 P(m ∪ B') = 1 - .70 = .30

Two pair: Two cards with one denomination, two with another, and one with a third. Example: ♣️, 3♥️, Q♠️, Q♥️, J♠️

Answer: 0.0475 13 ranks, choose 2 = c(12,2) 4 ranks, choose 2 = c(4,2) 44 ranks, choose 1 = c(44,1) 52 cards total, choose any 5 = c(52,5) [ c(12,2) • c(4,2) • c(44,1) ] / c(52,5)

Suppose that it snows in Greenland an average of once every 30 days, and when it does, glaciers have a 20% chance of growing. When it does not snow in Greenland, glaciers have only a 7% chance of growing. What is the probability that it is snowing in Greenland when glaciers are growing?

Answer: 0.0897 Snows: 1/30 x 20% = 20/30% Doesn't snow: 29/30 x 7% = 203/30% 20/30 + 203/30 = 223/30% (20/30) / (223/30) = 20/30 x 30/223 = 20/223 = .0897

There is a 60% chance of rain today and a 70% chance of rain tomorrow. Assume that the event that it rains today is independent of the event that it rains tomorrow. What is the probability that there will be no rain today or tomorrow?

Answer: 0.12 70% rain today = 1 - .70 = 30% no rain today 60% rain tomorrow = 1 - 60% = 40% no rain tomorrow Probability of no rain = 30% x 40% = 12% >>> 0.12 no rain today or tomorrow

Professor Frank Nabarro insists that all senior physics majors take his notorious physics aptitude test. The test is so tough that anyone not going on to a career in physics has no hope of passing, whereas 55% of the seniors who do go on to a career in physics still fail the test. Further, 90% of all senior physics majors in fact go on to a career in physics. Assuming that you fail the test, what is the probability that you will not go on to a career in physics?

Answer: 0.1681

The mean batting average in a certain baseball league is about 0.240. Supposing that batting averages are normally distributed, that the standard deviation in the averages is 0.05, and that there are 260 batters, what is the expected number of batters with an average of at least 0.4?†

Answer: 0.178 x̅= 0.24 σ= 0.05 Z score: [(X - x̅) /σ] <<< [(value - mean)/ S.T.D.] P(x ≥ 0.4) = p(z ≥ (0.4 - 0.24)/0.05) P(z ≥ 3.2) = 0.5 - 0.4993 0.000687026 x 260 >>>0.1786

According to a certain news poll, 64% agreed that it should be the government's responsibility to provide a decent standard of living for the elderly, and 36% agreed that it would be a good idea to invest part of their Social Security taxes on their own. If agreement with one of these propositions is independent of agreement with the other, what is the probability that a person agreed with both propositions?

Answer: 0.23 P(A) = 0.64 P(B) = 0.36 Formula: P(A ∩ B) = P(A) • P(B) P(A ∩ B) = 0.64 x 0.36 = 0.23

Your company issues flight insurance. You charge $5 and in the event of a plane crash, you will pay out $3 million to the victim or his or her family. In a certain year, the probability of a plane crashing on a single trip was .00000140. If ten people per flight buy insurance from you, what is your approximate probability of losing money over the course of 5 million flights?

Answer: 0.2451 N= 5,000,000 P= 0.0000014 x̅= 7 σ²= √(7)(0.9999986) σ= 2.64575 5 x 10 people = $50/flight ($50)(5,000,000) = $250,000,000 income Loss= $3 million x 10 = $30,000,000 loss >>> 250,000,000/30,000,000 = 8.33 crashes (P ≥ 8.333) >>> P(y ≥ 8.333 +/- 0.5) >>> P(y ≥ 8.833) P(2 ≥ ((8.83 - 7)/2.64575)) = P(z ≥ 0.6929353521) >>> P(z ≥ 0.693 - 0.5) >>> 0.2569029792 >>> 0.2451

If you roll a die 100 times, what is the approximate probability that you will roll between 10 and 16 ones, inclusive?

Answer: 0.45 N= 100 P= 1/6 Q= 1 - 1/6 = 5/6 x̅= n x p = 100 x 1/6 = 16.6667 σ²= √npq = √(100)(1/6)(5/6) σ= 3.7268 P(10 ≤ x ≤ 15) = P((9.5 ≤ y ≤ 16.5) P((9.5-16.5)/3.7268 ≤ z ≤ (15.5-16.6667)/3.7268)) >> P(-1.2923017 ≤ z ≤ -0.04473) >>> normalcdf: Lower: -1.923 Upper: -0.4473 x̅= 0 σ= 1 >>>>>= .45

According to a study of 200 samples of ground meats, the probability that one of the samples was contaminated by salmonella was .35. The probability that a salmonella-contaminated sample was contaminated by a strain resistant to at least one antibiotic was .76, and the probability that a salmonella-contaminated sample was contaminated by a strain resistant to at least three antibiotics was .55. Find the probability that a ground meat sample that was contaminated by an antibiotic-resistant strain was contaminated by a strain resistant to at least three antibiotics.

Answer: 0.724 Find: antibiotic - resistant strain contaminated by a strain resistant to at least 3 antibiotics Salmonella = .35 Strain resistant = .76 St 3 antibiotics = .55 >> 3 antibiotics/strain resistant = .55/.76 = .724

X has a normal distribution with the given mean and standard deviation. Find the indicated probability. μ = 59, σ = 10, find P(34 ≤ X ≤ 74)

Answer: 0.9270 P((34-59)/10 ≤ z ≤ (74-59)/10)) >>> (-2.5 ≤ z ≤ 1.5) Normal cdf: Low= -2.5 High= 1.5 x̅= 0 σ= 1 Answer= .927 Or Normal cdf: Low= 34 High= 74 x̅= 59 σ= 10 Answer: .927

❗️ Suppose that a monkey is seated at a computer keyboard and randomly strikes the 26 letter keys and the space bar. Find the probability that its first 39 characters (including spaces) will be "to be or not to be that is the question."

Answer: 1/(27^39) N(s) = 27^39 N(e) = 1 P(E) = N(e)/N(s) = 1 / (27^39)

An experiment is given together with an event. Find the (modeled) probability of each event, assuming that the coins are distinguishable and fair, and that what is observed are the faces uppermost. Three coins are tossed; the result is at most one head.

Answer: 1/2 S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} N(S) = 8 E = {HTT, THT, TTH, TTT} N(E) = 4 Formula: P(E) = N(S) / N(E) >>> 4/8 = 1/2

A poker hand consists of five cards from a standard deck of 52. Find the number of different poker hands of the specified type. Straight (five cards of consecutive denominations: A, 2, 3, 4, 5 up through 10, J, Q, K, A, not all of the same suit) (Note that the ace counts either as a 1 or as the denomination above king.)

Answer: 10200 hands 10 x 4^5 = 10240 *There are 40 possible straight flushes* 10240 - 40 = 10200 hands

❗️A bag contains 2 red marbles, 4 green ones, one lavender one, 6 yellows, and 4 orange marbles. How many sets of four marbles include all the red ones?

Answer: 105 sets Combination problem: 2r + 4g + 1L + 6y + 4o = 17 marbles total Four marbles include red: __ __ __ __ You have 2 red marbles: 1r 1r __ __ There are 15 marbles other than red Problem set up: c(2,2) x c(15,2) = 105 sets

How many five-letter sequences are possible that use the letters b, p, g, f, x once each?

Answer: 120 sequences Factorial! >>> 5! >>> 5 x 4 x 3 x 2 x 1 = 120

Recall from Example 1 that whenever Suzan sees a bag of marbles, she grabs a handful at random. She has seen a bag containing four red marbles, three green ones, four white ones, and two purple ones. She grabs eight of them. Find the probability of the following event, expressing it as a fraction in lowest terms. She has at least one green one.

Answer: 138/143 N(e) = c(3,1) • c(10,7) = 360 N(e) = c(3,2) • c(10,6)= 630 N(e) = c(3,3) • c(10,5)= 252 360 + 630 + 252 = 1242 P(E) = N(e)/N(s) = 1242/1287 = 138/143

Recall from Example 1 that whenever Suzan sees a bag of marbles, she grabs a handful at random. She has seen a bag containing three red marbles, two green ones, four white ones, and three purple ones. She grabs eight of them. Find the probability of the following event, expressing it as a fraction in lowest terms. She has all the red ones.

Answer: 14/55 N(s) = c(12,8) = 495 N(e) = c(3,3) • c(9,5)= 126 P(E) = N(e)/N(s) = 126/495 = 14/55

❗️ In a New York State daily lottery game, a sequence of 3 digits (not necessarily different) in the range 0-9 are selected at random. Find the probability that three are different.

Answer: 18/25 N(E)/N(S) = P(10,3)/10^3 = 720/1000 = 18/25

An experiment is given together with an event. Find the (modeled) probability of each event, assuming that the dice are distinguishable and fair, and that what is observed are the numbers uppermost. Two dice are rolled; the numbers add to 11.

Answer: 2/36 1/6 + 1/6 • 1/6 = 2/6 • 1/6 = 2/36

Recall from Example 1 that whenever Suzan sees a bag of marbles, she grabs a handful at random. She has seen a bag containing three red marbles, four green ones, three white ones, and one purple one. She grabs seven of them. Find the probability of the following event, expressing it as a fraction in lowest terms. She does not have all the red ones.

Answer: 26/33 N(s) = c(11,7) = 330 N(e) = c(3,3) • c(8,4)= 70 P(E) = N(e)/N(s) = 70/330 = 7/33 P(E') = 1 - 7/33 = 26/33

How many "ordered" lists are there of eleven items chosen from from thirteen?

Answer: 3113510400 ordered lists Permutation>> P(13,11)

A bag contains two red marbles, five green ones, one lavender one, two yellows, and two orange marbles. How many sets of five marbles include either the lavender one or exactly one yellow one but not both colors?

Answer: 378 sets C(3,1) x c(9,4) = 378

A bag contains 3 red marbles, 2 green ones, 1 lavender one, 4 yellows, and 2 orange marbles. How many sets of four marbles include one of each color other than lavender?

Answer: 48 sets Factorial problem: 3r x 2g x 4y x 2o = 48 Lavender is not included

How many "ordered" sequences are possible that contain eleven objects from fifteen?

Answer: 54486432000 ordered sequences Combination>>> P(15,11)

A poker hand consists of five cards from a standard deck of 52. Find the number of different poker hands of the specified type. Three of a kind (three of one denomination, one of another denomination, and one of a third)

Answer: 54912 hands S1: c(13,1) = 13 S2: c(4,3) = 4 S3: c(12,2) = 66 S4: c(4,1) = 4 S5: c(13,1) = 4 13 x 4 x 66 x 4 x 4 = 54912 hands

How many six-letter sequences are possible that use the letters d, u, d, d, u, k?

Answer: 60 sequences 3ds 2us 1k ; 6 choices C(5,3) = you have 5 choices left and three ds to choose from C(2,2) = you have 2 choices left and 3 us to choose from 6 choices x c(5,3) x c(2,2) = 60

Recall from Example 1 that whenever Suzan sees a bag of marbles, she grabs a handful at random. She has seen a bag containing four red marbles, three green ones, five white ones, and two purple ones. She grabs five of them. Find the probability of the following event, expressing it as a fraction in lowest terms. She has two red ones and one of each of the other colors.

Answer: 90/1001 N(s) = c(14,5) = 2002 N(e) = c(4,2) • c(3,1) • c(5,1) • c(2,1) = 126 P(E) = N(e)/N(s) = 180/2002 = 90/1001

Calculate the expected value E(X) of the given random variable X. 30 darts are thrown at a dartboard. The probability of hitting a bull's-eye is .1. Let X be the number of bull's-eyes hit. E(x) =

Answer: E(x) = 3 N= 30 P= 0.1 E(x) = n x p E(x) = 30 x 0.1 = 3

Assume that on a standardized test of 100 questions, a person has a probability of 80% of answering any particular question correctly. Find the probability of answering between 72 and 82 questions, inclusive, correctly. P(72 ≤ X ≤ 82) =

Answer: P(72 ≤ X ≤ 82) = 0.7088 P= 0.8 Q= 1 - 0.8 = 0.2 c(100,72) (0.8)^72 (0.2)^28 c(100,73) (0.8)^73 (0.2)^27 c(100,74) (0.8)^74 (0.2)^26 c(100,75) (0.8)^75 (0.2)^25 c(100,76) (0.8)^76 (0.2)^24 c(100,77) (0.8)^77 (0.2)^23 c(100,78) (0.8)^78 (0.2)^22 c(100,79) (0.8)^79 (0.2)^21 c(100,80) (0.8)^80 (0.2)^20 c(100,81) (0.8)^81 (0.2)^19 c(100,82) (0.8)^82 (0.2)^18 Add all together = 0.708790293 = 0.7088

P(A) = .79. Find P(A').

Answer: P(A)= .21 Formula: P(A') = 1 - P(A) A' = 1 - .79 A' = .21

P(A ∪ B) = .9, P(B) = .7, P(A ∩ B) = .6. Find P(A).

Answer: P(A)=.8 Formula: P(A ∪ B) = A + B - (A ∩ B) .9 = A + .7 - (.6) 1.5 = A + .7 .8 = A

P(A | B) = .1, P(B) = .5, P(A | B') = .3. Find P(B | A).

Answer: P(B | A) = 0.25 Formula: P(B | A) = [ P(A | B) • P(B) ] / [ P(A | B) • P(B) + P(A | B') P(B') P(B') = 1 - P(B) >>> 1 - 0.5 = 0.5 >> (0.1)(0.5) / (0.1)(0.5)+(0.3)(0.5) P(B | A) = 0.25

In a large on-the-job training program, half of the participants are female and half are male. In a random sample of six participants, what is the probability that an investigator will draw at least two males?† P(X ≥ 2) =

Answer: P(X ≥ 1) = 0.9375 N= 4 P= 1/2 Q= 1 - P = 1 - 1/2 = 1/2 c(4,4) (0.5^4) (0.5)^0 c(4,3) (0.5^3) (0.5)^1 c(4, 2) (0.5^2) (0.5)^2 c(4,3) (0.5^1) (0.5)^3 Add all together = 0.9475

You are performing 6 independent Bernoulli trials with p = 0.3 and q = 0.7. Calculate the probability of the stated outcome. Check your answer using technology. Two successes.

Answer: P(x=2) = 0.3241 c(6,2) = 6 Independent Bernoulli trials and 2 successes N=6 P= 0.3 Q= 0.7 Formula: n x p^n • q^x-n P(x=2) = c(6,2) (0.3)^2 (0.7)^4

You are performing 3 independent Bernoulli trials with p = 0.2 and q = 0.8. Calculate the probability of the stated outcome. Check your answer using technology. No failures.

Answer: P(x=3) = 0.008 N= 3 P= 0.2 Q= 0.8 c(3,3) = 3 trials and 3 wins Formula: n x p^n • q^x-n P(x=2) = c(3,3) (0.2)^3 (0.8)^0 = 0.008

Calculate the expected value E(X) of the given random variable X. X is the number of tails that come up when a coin is tossed 62 times.

E(X) = 31 62/2 = 31

Calculate the expected value E(X) for the given random variable X. [This exercise assumes familiarity with counting arguments and probability Select thirteen cards without replacement from a standard deck of 52, and let X be the number of red cards you draw.

E(x) = 6.5 (13 cards/ 52 total) (26 red cards) = 6.5