Stats
What is the chance of getting either 1, 2, 3, 4, 5, or 6 on the next roll?
100%. The outcome must be one of these numbers.
Portfolio return. A portfolio's value increases by 18% during a financial boom and by 9% during normal times. It decreases by 12% during a recession. What is the expected return on this portfolio if each scenario is equally likely?
5% increase in value.
Predisposition for thrombosis. A genetic test is used to determine if people have a predisposition for thrombosis, which is the formation of a blood clot inside a blood vessel that obstructs the flow of blood through the circulatory system. It is believed that 3% of people actually have this predisposition. The genetic test is 99% accurate if a person actually has the predisposition, meaning that the probability of a positive test result when a person actually has the predisposition is 0.99. The test is 98% accurate if a person does not have the predisposition. What is the probability that a randomly selected person who tests positive for the predisposition by the test actually has the predisposition?
=(0.03*0.99)/ [(.03*.99)+(.97*.02)] =0.604887 = 60.5%
The outcome of interest is whether there is a sporting event (call this A1), and the condition is that the lot is full (B). Let A2 represent an academic event and A3 represent there being no event on campus. Then the given probabilities can be written as: P(A1) = 0.2 ,,, P(A2) = 0.35 ,,, P(A3) = 0.45 P (B|A1) = 0.7 ,,, P (B|A2) = 0.25 ,,, P (B|A3) = 0.05 Bayes' Theorem can be used to compute the probability of a sporting event (A1) under the condition that the parking lot is full (B)
Based on the information that the garage is full, there is a 56% probability that a sporting event is being held on campus that evening.
In the loans data set describing 10,000 loans, 1495 loans were from joint applications (e.g. a couple applied together), 4789 applicants had a mortgage, and 950 had both of these characteristics. Create a Venn diagram for this setup.
Both the counts and corresponding probabilities (e.g. 3839/10000 = 0.384) are shown. Notice that the number of loans represented in the left circle corresponds to 3839 + 950 = 4789, and the number represented in the right circle is 950 + 545 = 1495.
The people of Boston self-selected whether or not to be inoculated. (a) Is this study observational or was this an experiment? (b) Can we infer any causal connection using these data? (c) What are some potential confounding variables that might influence whether someone lived or died and also affect whether that person was inoculated
Brief answers: (a) Observational. (b) No, we cannot infer causation from this observational study. (c) Accessibility to the latest and best medical care. There are other valid answers for part (c).
Answer: Books on a bookshelf
(a) (28/95) * (59/94) = .185 (b) (72/95) * (28/94) = .2258 (c) (72/95) * (28/95) = .2234 (d) They are similar because there is a probability that the fiction book is hardcover, and it therefore impactsthe probability that the second draw is a hardcover book.
Three US adults are randomly selected. The probability a single adult is between 180 and 185 cm is 0.1157.60 (a) What is the probability that all three are between 180 and 185 cm tall? (b) What is the probability that none are between 180 and 185 cm?
(a) 0.1157 × 0.1157 × 0.1157 = 0.0015. (b) (1 − 0.1157)3 = 0.692
Educational attainment of couples. The table below shows the distribution of education level attained by US residents by gender based on data collected in the 2010 American Community Survey. (a) What is the probability that a randomly chosen man has at least a Bachelor's degree? (b) What is the probability that a randomly chosen woman has at least a Bachelor's degree? (c) What is the probability that a man and a woman getting married both have at least a Bachelor's degree? Note any assumptions you must make to answer this question. (d) If you made an assumption in part (c), do you think it was reasonable? If you didn't make an assumption, double check your earlier answer and then return to this part.
(a) 0.16 + 0.09 = 0.25. (b) 0.17 + 0.09 = 0.26. (c) Assuming that the education level of the hus- band and wife are independent: 0.25 × 0.26 = 0.065. You might also notice we actually made a second assumption: that the decision to get married is unrelated to education level. (d) The husband/wife in- dependence assumption is probably not reasonable, because people often marry another person with a comparable level of education. We will leave it to you to think about whether the second assumption noted in part (c) is reasonable.
Marbles in an urn. Imagine you have an urn containing 5 red, 3 blue, and 2 orange marbles in it. (a) What is the probability that the first marble you draw is blue? (b) Suppose you drew a blue marble in the first draw. If drawing with replacement, what is the probability of drawing a blue marble in the second draw? (c) Suppose you instead drew an orange marble in the first draw. If drawing with replacement, what is the probability of drawing a blue marble in the second draw? (d) If drawing with replacement, what is the probability of drawing two blue marbles in a row? (e) When drawing with replacement, are the draws independent? Explain.
(a) 0.3. (b) 0.3. (c) 0.3. (d) 0.3 × 0.3 = 0.09. (e) Yes, the population that is being sampled from is identical in each draw.
Coin flips. If you flip a fair coin 10 times, what is the probability of (a) getting all tails? (b) getting all heads? (c) getting at least one tails?
(a) 0.5^10 = 0.00098. (b) 0.5^10 = 0.00098. (c) P(at least one tails) = 1 − P(no tails) = 1 − (0.5^10) ≈ 1 − 0.001 = 0.999.
Let X and Y represent the outcomes of rolling two dice. (a) What is the probability that the first die, X, is 1? (b) What is the probability that both X and Y are 1? (c) Use the formula for conditional probability to compute P (Y = 1 | X = 1). (d) What is P (Y = 1)? Is this different from the answer from part (c)? Explain.
(a) 1/6. (b) 1/36. (c) P (Y = 1 and X = 1)/P(X = 1) = (1/36)/(1/6) = 1/6. (d) The probability is the same as in P(X= 1) 1/6 part (c): P(Y = 1) = 1/6. The probability that Y = 1 was unchanged by knowledge about X, which makes sense as X and Y are independent.
Below are four versions of the same game. Your archnemesis gets to pick the version of the game, and then you get to choose how many times to flip a coin: 10 times or 100 times. -Identify how many coin flips you should choose for each version of the game. -It costs $1 to play each game. (a) If the proportion of heads is larger than 0.60, you win $1. (b) If the proportion of heads is larger than 0.40, you win $1. (c) If the proportion of heads is between 0.40 and 0.60, you win $1. (d) If the proportion of heads is smaller than 0.30, you win $1.
(a) 10 tosses. Fewer tosses mean more variabil- ity in the sample fraction of heads, meaning there's a better chance of getting at least 60% heads. (b) 100 tosses. More flips means the observed proportion of heads would often be closer to the average, 0.50, and therefore also above 0.40. (c) 100 tosses. With more flips, the observed proportion of heads would often be closer to the average, 0.50. (d) 10 tosses. Fewer flips would increase variability in the fraction of tosses that are heads.
The bookstore also offers a chemistry textbook for $159 and a book supplement for $41. From past experience, they know about 25% of chemistry students just buy the textbook while 60% buy both the textbook and supplement. (a) What proportion of students don't buy either book? Assume no students buy the supplement without the textbook. (b) Let Y represent the revenue from a single student. Write out the probability distribution of Y , i.e. a table for each outcome and its associated probability. (c) Compute the expected revenue from a single chemistry student. (d) Find the standard deviation to describe the variability associated with the revenue from a single student.
(a) 100% - 25% - 60% = 15% of students do not buy any books for the class. (b) is represented by the first two lines in the table below. The expectation for part (c) is given as the total on the line yi ×P(Y = yi). The result of part (d) is the square-root of the variance listed on in the total on the last line: σ = sqrt(Var(Y ) = $69.28.)
College smokers. At a university, 13% of students smoke. (a) Calculate the expected number of smokers in a random sample of 100 students from this university. (b) The university gym opens at 9 am on Saturday mornings. One Saturday morning at 8:55 am there are 27 students outside the gym waiting for it to open. Should you use the same approach from part (a) to calculate the expected number of smokers among these 27 students?
(a) 13. (b) No, these 27 students are not a ran- dom sample from the university's student population. For example, it might be argued that the proportion of smokers among students who go to the gym at 9 am on a Saturday morning would be lower than the proportion of smokers in the university as a whole.
Roulette wheel. The game of roulette involves spinning a wheel with 38 slots: 18 red, 18 black, and 2 green. A ball is spun onto the wheel and will eventually land in a slot, where each slot has an equal chance of capturing the ball. (a) You watch a roulette wheel spin 3 consecutive times and the ball lands on a red slot each time. What is the probability that the ball will land on a red slot on the next spin? (b) You watch a roulette wheel spin 300 consecutive times and the ball lands on a red slot each time. What is the probability that the ball will land on a red slot on the next spin? (c) Are you equally confident of your answers to parts (a) and (b)? Why or why not?
(a) 18/36 = 0.4737 = 47% (b) 18/36 = 0.4737 = 47% (c)...yes if the roulette wheel is fair; the 300 consecutive red slot outcomes might suggest that the roulette wheel is not fair
Chips in a bag. Imagine you have a bag containing 5 red, 3 blue, and 2 orange chips. (a) Suppose you draw a chip and it is blue. If drawing without replacement, what is the probability the next is also blue? (b) Suppose you draw a chip and it is orange, and then you draw a second chip without replacement. What is the probability this second chip is blue? (c) If drawing without replacement, what is the probability of drawing two blue chips in a row? (d) When drawing without replacement, are the draws independent? Explain.
(a) 2/9 ≈ 0.22. (b) 3/9 ≈ 0.33. (c) 3/10 × 2/9 ≈ 0.067. (d) No, e.g. in this exercise, removing one chip meaningfully changes the probability of what might be drawn next.
Events A = {1, 2} and B = {4, 6} are shown. (a) Write out what A^c and B^c represent. (b) Compute P (A^c) and P (B^c). (c) Compute P (A) + P (A^c) and P (B) + P (B^c).
(a) A^c = {3, 4, 5, 6} and B^c = {1, 2, 3, 5}. (b) Noting that each outcome is disjoint, add the individual outcome probabilities to get P(A^c) = 2/3 and P(B^c) = 2/3. (c) A and A^c are disjoint, and the same is true of B and B^c. Therefore, P(A) + P(A^c) = 1 and P(B) + P(B^c) = 1.
Question: Burger preferences. A 2010 SurveyUSA poll asked 500 Los Angeles residents, "What is the best hamburger place in Southern California? Five Guys Burgers? In-N-Out Burger? Fat Burger? Tommy's Hamburgers? Umami Burger? Or somewhere else?" The distribution of responses by gender is shown below
(a) Are being female and liking Five Guys Burgers mutually exclusive? (b) What is the probability that a randomly chosen male likes In-N-Out the best? (c) What is the probability that a randomly chosen female likes In-N-Out the best? (d) What is the probability that a man and a woman who are dating both like In-N-Out the best? Note any assumption you make and evaluate whether you think that assumption is reasonable. (e) What is the probability that a randomly chosen person likes Umami best or that person is female?
Question: Health coverage, relative frequencies. The Behavioral Risk Factor Surveillance System (BRFSS) is an annual telephone survey designed to identify risk factors in the adult population and report emerging health trends. The following table displays the distribution of health status of respondents to this survey (excellent, very good, good, fair, poor) and whether or not they have health insurance.
(a) Are being in excellent health and having health coverage mutually exclusive? (b) What is the probability that a randomly chosen individual has excellent health? (c) What is the probability that a randomly chosen individual has excellent health given that he has health coverage? (d) What is the probability that a randomly chosen individual has excellent health given that he doesn't have health coverage? (e) Do having excellent health and having health coverage appear to be independent?
Question: Global warming. A Pew Research poll asked 1,306 Americans "From what you've read and heard, is there solid evidence that the average temperature on earth has been getting warmer over the past few decades, or not?". The table below shows the distribution of responses by party and ideology, where the counts have been replaced with relative frequencies.
(a) Are believing that the earth is warming and being a liberal Democrat mutually (b) What is the probability that a randomly chosen respondent believes the earth is warming or is a liberal Democrat? (c) What is the probability that a randomly chosen respondent believes the earth is warming given that he is a liberal Democrat? (d) What is the probability that a randomly chosen respondent believes the earth is warming given that he is a conservative Republican? (e) Does it appear that whether or not a respondent believes the earth is warming is independent of their party and ideology? Explain your reasoning. (f) What is the probability that a randomly chosen respondent is a moderate/liberal Republican given that he does not believe that the earth is warming?
Two books are assigned for a statistics class: a textbook and its corresponding study guide. The university bookstore determined 20% of enrolled students do not buy either book, 55% buy the textbook only, and 25% buy both books, and these percentages are relatively constant from one term to another. (a) If there are 100 students enrolled, how many books should the bookstore expect to sell to this class? (b) Would you be surprised if the bookstore sold slightly more or less than 105 books (c) The textbook costs $137 and the study guide $33. How much revenue should the bookstore expect from this class of 100 students? (d) What is the average revenue per student for this course? (e) here is the probability distribution
(a) Around 20 students will not buy either book (0 books total), about 55 will buy one book (55 books total), and approximately 25 will buy two books (totaling 50 books for these 25 students). The bookstore should expect to sell about 105 books for this class. (b) If they sell a little more or a little less, this should not be a surprise. (c) textbook only: $137*55 = $7535 textbook and study guide: ($137 + $33)*25 = $4250 total sales = $7535 + $4250 = $11,785 (However, there might be some sampling variability so the actual amount may differ by a little bit.) (d) The expected total revenue is $11,785, and there are 100 students. Therefore the expected revenue per student is $11, 785/100 = $117.85 (e) The possible outcomes of X are labeled with a corresponding lower case letter x and subscripts. For example, we write x1 = $0, x2 = $137, and x3 = $170, which occur with probabilities 0.20, 0.55, and 0.25. The distribution of X is summarized in Figure 3.19.
Hearts win. In a new card game, you start with a well-shuffled full deck and draw 3 cards without replacement. If you draw 3 hearts, you win $50. If you draw 3 black cards, you win $25. For any other draws, you win nothing. (a) Create a probability model for the amount you win at this game, and find the expected winnings. Also compute the standard deviation of this distribution. (b) If the game costs $5 to play, what would be the expected value and standard deviation of the net profit (or loss)? (Hint: profit = winnings − cost; X − 5) (c) If the game costs $5 to play, should you play this game? Explain.
(a) E(X) = 3.59. SD(X) = 9.64. (b) E(X) = -1.41. SD(X) = 9.64. (c) No, the expected net profit is negative, so on average you expect to lose money.
Question: Books on a bookshelf. The table below shows the distribution of books on a bookcase based on whether they are nonfiction or fiction and hardcover or paperback.
(a) Find the probability of drawing a hardcover book first then a paperback fiction book second when drawing without replacement. (b) Determine the probability of drawing a fiction book first and then a hardcover book second, when drawing without replacement. (c) Calculate the probability of the scenario in part (b), except this time complete the calculations under the scenario where the first book is placed back on the bookcase before randomly drawing the second book. (d) The final answers to parts (b) and (c) are very similar. Explain why this is the case.
Your department is holding a raffle. They sell 30 tickets and offer seven prizes. (a) They place the tickets in a hat and draw one for each prize. The tickets are sampled without replacement, i.e. the selected tickets are not placed back in the hat. What is the probability of winning a prize if you buy one ticket? (b) What if the tickets are sampled with replacement?
(a) First determine the probability of not winning. The tickets are sampled without replacement, which means the probability you do not win on the first draw is 29/30, 28/29 for the second, ..., and 23/24 for the seventh. The probability you win no prize is the product of these separate probabilities: 23/30. That is, the probability of winning a prize is 1 − 23/30 = 7/30 = 0.233. (b) When the tickets are sampled with replacement, there are seven independent draws. Again we first find the probability of not winning a prize: (29/30)7 = 0.789. Thus, the probability of winning (at least) one prize when drawing with replacement is 0.211.
Find the following probabilities for rolling two dice: (a) The sum of the dice is not 6. (b) The sum is at least 4. That is, determine the probability of the event B = {4, 5, ..., 12}. (c) The sum is no more than 10. That is, determine the probability of the event D = {2, 3, ..., 10}.
(a) First find P (6) = 5/36, then use the complement: P (not 6) = 1 − P (6) = 31/36. (b) First find the complement, which requires much less effort: P (2 or 3) = 1/36 + 2/36 = 1/12. Then calculate P(B) = 1 − P(Bc) = 1 − 1/12 = 11/12.(c) As before, finding the complement is the clever way to determine P(D). First find P(Dc) = P(11 or 12) = 2/36 + 1/36 = 1/12. Then calculate P(D) = 1 − P(Dc) = 11/12.
(a) If A and B are disjoint, describe why this implies P(A and B) = 0. (b) Using part (a), verify that the General Addition Rule simplifies to the simpler Addition Rule for disjoint events if A and B are disjoint
(a) If A and B are disjoint, A and B can never occur simultaneously. (b) If A and B are disjoint, then the last P(A and B) term of in the General Addition Rule formula is 0 (see part (a)) and we are left with the Addition Rule for disjoint events.
Disjoint vs. independent. In parts (a) and (b), identify whether the events are disjoint, independent, or neither (events cannot be both disjoint and independent). (a) You and a randomly selected student from your class both earn A's in this course. (b) You and your class study partner both earn A's in this course. (c) If two events can occur at the same time, must they be dependent?
(a) If the class is not graded on a curve, they are independent. If graded on a curve, then neither independent nor disjoint - unless the instructor will only give one A, which is a situation we will ignore in parts (b) and (c). (b) They are probably not in- dependent: if you study together, your study habits would be related, which suggests your course perfor- mances are also related. (c) No. See the answer to part (a) when the course is not graded on a curve. More generally: if two things are unrelated (independent), then one occurring *does not preclude* the other from occurring.
3.29 (a) Write out the following statement in conditional probability notation: "The probability that the ML prediction was correct, if the photo was about fashion". Here the condition is now based on the photo's truth status, not the ML algorithm. (b) Determine the probability from part (a)
(a) If the photo is about fashion and the ML algorithm prediction was correct, then the ML algorithm my have a value of pred fashion: P (mach learn is pred fashion | truth is fashion) (b) The equation for conditional probability indicates we should first findP (mach learn is pred fashion and truth is fashion) = 0.1081 and P (truth is fashion) = 0.1696. Then the ratio represents the conditional probability: 0.1081/0.1696 = 0.6374.
The birthday problem. Suppose we pick three people at random. For each of the following questions, ignore the special case where someone might be born on February 29th, and assume that births are evenly distributed throughout the year. (a) What is the probability that the first two people share a birthday? (b) What is the probability that at least two people share a birthday?
(a) If there are 2 people, the chance that they do not have the same birthday is 364/365, so the chance that they do have the same birthday is 1- 364/365 = 1/365≈0.28%. (b) If there are 3 people, you and 2 others, the chance that neither of the other two shares your specific birthday is 364/365 * 364/365 and so the chance that no one else shares your birthday is 1- 364/365 * 364/365≈ 0.55%. However, the other two might have the same birthday, not equal to yours. The chance that all 3 people have different birthdays is 365/365 * 364/365 * 363/365 hence, the probability that not all three birthdays are distinct (i.e. at least two share the same birthday) is 1-365/365 * 364/365 * 363/365≈ 0.82%
Leonard has invested $6000 in Caterpillar Inc (stock ticker: CAT) and $2000 in Exxon Mobil Corp (XOM). (a) If X represents the change in Caterpillar's stock next month and Y represents the change in Exxon Mobil's stock next month, write an equation that describes how much money will be made or lost in Leonard's stocks for the month. (b) Caterpillar stock has recently been rising at 2.0% and Exxon Mobil's at 0.2% per month, respectively. Compute the expected change in Leonard's stock portfolio for next month (c) You should have found that Leonard expects a positive gain. However, would you be surprised if he actually had a loss this month?
(a) Leonard's net gain = $6000x + $2000y (b) E($6000×X +$2000×Y) = $6000×0.020+$2000×0.002 = $124 (c) No. While stocks tend to rise over time, they are often volatile in the short term.
Joint and conditional probabilities. P(A) = 0.3, P(B) = 0.7 (a) Can you compute P(A and B) if you only know P(A) and P(B)? (b) Assuming that events A and B arise from independent random processes, i. what is P(A and B)? ii. what is P(A or B)? iii. what is P(A|B)? (c) If we are given that P(A and B) = 0.1, are the random variables giving rise to events A and B indepen- dent? (d) If we are given that P(A and B) = 0.1, what is P(A|B)?
(a) No, but we could if A and B are independent. (b-i) 0.21. (b-ii) 0.79. (b-iii) 0.3. (c) No, because 0.1 does not 0.21, where 0.21 was the value computed under independence from part (a). (d) 0.143.
Poverty and language. The American Community Survey is an ongoing survey that provides data every year to give communities the current information they need to plan investments and services. The 2010 American Community Survey estimates that 14.6% of Americans live below the poverty line, 20.7% speak a language other than English (foreign language) at home, and 4.2% fall into both categories. (a) Are living below the poverty line and speaking a foreign language at home disjoint? (b) Draw a Venn diagram summarizing the variables and their associated probabilities. (c) What percent of Americans live below the poverty line and only speak English at home? (d) What percent of Americans live below the poverty line or speak a foreign language at home? (e) What percent of Americans live above the poverty line and only speak English at home? (f) Is the event that someone lives below the poverty line independent of the event that the person speaks a foreign language at home?
(a) No, there are people who both live below the poverty line and do not speak english at home (c) P(Below Poverty, yes english at home) = 14.6% - 4.2% = 10.4% (d) P(Below Poverty or no english at home) = 14.6% + 20.7% - 4.2% = 31.1% (e) P(Above Poverty, yes english at home) = 1-0.311 = 0.689 = 68.9% (f) P(below poverty and no english at home) = 0.146*0.207 = 0.03033 = 3.0% not equal t to reported 4.2%, therefore they are dependent
Swing voters. A Pew Research survey asked 2,373 randomly sampled registered voters their political affiliation (Republican, Democrat, or Independent) and whether or not they identify as swing voters. 35% of respondents identified as Independent, 23% identified as swing voters, and 11% identified as both. (a) Are being Independent and being a swing voter disjoint, i.e. mutually exclusive? (b) Draw a Venn diagram summarizing the variables and their associated probabilities. (c) What percent of voters are Independent but not swing voters? (d) What percent of voters are Independent or swing voters? (e) What percent of voters are neither Independent nor swing voters? (f) Is the event that someone is a swing voter independent of the event that someone is a political Indepen- dent?
(a) No, there are voters who are both indepen- dent and swing voters. (c) Each Independent voter is either a swing voter or not. Since 35% of voters are Independents and 11% are both Independent and swing voters, the other 24% must not be swing voters. (d) P(Independent or Swing) = P(Independent) + P(Swing) - P(Independent and Swing)= 35% + 23% - 11% = 47% (e) P(Neither Independent nor Swing) = 1-P(Independent or Swing) = 1-0.47 = 0.54 = 53% (f) P(Independent) × P(swing) = 0.35×0.23 = 0.08, which does not equal P(Independent and swing) = 0.11, so the events are dependent.
Answer: Burger preferences
(a) No. There are 6 females who like Five Guys Burgers. (b) 162/248 = 0.65. (c) 181/252 = 0.72. (d) Under the assumption of a dating choices be- ing independent of hamburger preference, which on the surface seems reasonable: 0.65 × 0.72 = 0.468. (e) (252 + 6 − 1)/500 = 0.514.
(a) Using Figure 3.2 as a reference, what outcomes are represented by event D? (b) Are events B and D disjoint? (c) Are events A and D disjoint?
(a) Outcomes 2 and 3. (b) Yes, events B and D are disjoint because they share no outcomes. (c) The events A and D share an outcome in common, 2, and so are not disjoint.
School absences. Data collected at elementary schools in DeKalb County, GA suggest that each year roughly 25% of students miss exactly one day of school, 15% miss 2 days, and 28% miss 3 or more days due to sickness.24 (a) What is the probability that a student chosen at random doesn't miss any days of school due to sickness this year? (b) What is the probability that a student chosen at random misses no more than one day? (c) What is the probability that a student chosen at random misses at least one day? (d) If a parent has two kids at a DeKalb County elementary school, what is the probability that neither kid will miss any school? Note any assumption you must make to answer this question. (e) If a parent has two kids at a DeKalb County elementary school, what is the probability that both kids will miss some school, i.e. at least one day? Note any assumption you make. (f) If you made an assumption in part (d) or (e), do you think it was reasonable? If you didn't make any assumptions, double check your earlier answers.
(a) P(0 sick day) = 1-P(1 sick day) - P(2 sick days) - P(3+ sick day) = 1-0.25-0.15-0.28 = 0.32 = 32% (b) P(1 sick day) + P(0 sick days)= 0.25 + 0.32 = 0.57 (c) P(at least 1 sick day) = 1-P(0 sick days) = 1-0.32 = 0.68 = 68% (d) assuming independent events, P(neither miss any) = P(0 sick days)*P(0 sick days) = 0.32^2 = 0.1024 = 10% (e) assuming independent events, P(at lease 1 sick day)*P(at least 1 sick day) = 0.68^2 = 0.4624 = 46% (f) no, not reasonable at all, but without additional information that was the necessary assumption
Socks in a drawer. In your sock drawer you have 4 blue, 5 gray, and 3 black socks. Half asleep one morning you grab 2 socks at random and put them on. Find the probability you end up wearing (a) 2 blue socks (b) no gray socks (c) at least 1 black sock (d) a green sock (e) matching socks
(a) P(blue and blue) = (4/12)*(3/11) = 0.0909 = 9% (b) P(no grey) = ((12-5)/12)*((11-5)/11) (c) P(at least one black) = P(black and not black) + P(not black and black) + p(black and black) = (3/12)*(9/11) + (9/12)*(3/11) + (3/12)*(2/11) = 0.454545 = 45% OR P(at least one black) = 1-P(no black) = 1-((9/12)*(8/11))= 0.454545 = 45% (d) 0 (e) 2 blue OR 2 gray OR 2 black A = P(2 blue) = 4/12 * 3/11 = .0909 B = P(2 gray) = 5/12 * 4/11 = .1515 C = P(2 black) = 3/12 * 2/11 = .04545 P (A U B U C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P(A ∩ B ∩ C) = .0909 + .1515 + .04545 - (.0909 * .1515) - (.0909 * .04545) - (.1515 * .04545) + (.0909 * .1515 * .04545) = .2637
The smallpox data set provides a sample of 6,224 individuals from the year 1721 who were exposed to smallpox in Boston. Doctors at the time believed that inoculation, which involves exposing a person to the disease in a controlled form, could reduce the likelihood of death. Each case represents one person with two variables: inoculated and result. The variable inoculated takes two levels: yes or no, indicating whether the person was inoculated or not. The variable result has outcomes lived or died. (a) Write out, in formal notation, the probability a randomly selected person who was not inoculated died from smallpox, and find this probability. (b) Determine the probability that an inoculated person died from smallpox. How does this result compare with the result of (a)
(a) P(result = died | inoculated = no) =P(result = died and inoculated = no)/P(inoculated = no) = 0.1356/0.9608 = 0.1411 (b) P(result = died | inoculated = yes) = P(result = died and inoculated = yes)/= died and inoculat = 0.0010/0.0392= 0.0255 The death rate for individuals who were inoculated is only about 1 in 40 while the death rate is about 1 in 7 for those who were not inoculated.
Dice rolls. If you roll a pair of fair dice, what is the probability of (a) getting a sum of 1? (b) getting a sum of 5? (c) getting a sum of 12?
(a) P(sum of 1) = 0 (b) P(sum of 5) = P(1 & 4) + P(4 & 1) + P(2 & 3) = P(3 & 2) = [(1/6)*(1/6)]*4 = 0.111=11% (c) P(sum of 12) = P(6 & 6) = (1/6)*(1/6) = 1/36
Guessing on an exam. In a multiple choice exam, there are 5 questions and 4 choices for each question (a, b, c, d). Nancy has not studied for the exam at all and decides to randomly guess the answers. What is the probability that: (a) the first question she gets right is the 5th question? (b) she gets all of the questions right? (c) she gets at least one question right?
(a) P(wrong) * P(wrong) * P(wrong) * P(wrong) * P(right) = (3/4)*(3/4)*(3/4)*(3/4)*(1/4) = 0.0791 = 8% (b) P(right)^5 = 0.25^5 = 9.765e-4 = 0.1% (c) P(at least one right) = 1-p(all wrong) = 1 - 0.75^5 = 0.76 = 76%
Elena is selling a TV at a cash auction and also intends to buy a toaster oven in the auction. (a) If X represents the profit for selling the TV and Y represents the cost of the toaster oven, write an equation that represents the net change in Elena's cash (b) Based on past auctions, Elena figures she should expect to make about $175 on the TV and pay about $23 for the toaster oven. In total, how much should she expect to make or spend? (c) Would you be surprised if Elena didn't make exactly $152?
(a) She will make X dollars on the TV but spend Y dollars on the toaster oven: X − Y . (b) E(X −Y) = E(X)−E(Y) = 175−23 = $152. She should expect to make about $152 (c) No, since there is probably some variability; and auction prices will vary depending on the quality of the merchandise and the interest of the attendees.
About 9% of people are left-handed. Suppose 5 people are selected at random. (a) What is the probability that all are right-handed? (b) What is the probability that all are left-handed? (c) What is the probability that not all of the people are right-handed?
(a) Since each are independent, we apply the Multiplication Rule for independent processes: P(all five are RH) = P(first = RH, second = RH, ..., fifth = RH)= P (first = RH) × P (second = RH) × · · · × P (fifth = RH) = 0.91×0.91×0.91×0.91×0.91 = 0.624 (b) Using the same reasoning as in (a), 0.09 × 0.09 × 0.09 × 0.09 × 0.09 = 0.0000059 (c) Use the complement, P (all five are RH), to answer this question:P(not all RH) = 1−P(all RH) = 1−0.624 = 0.376
(a) Compute P(Dc) = P(rolling a 1, 4, 5, or 6). (b) What is P(D) + P(Dc)?
(a) The outcomes are disjoint and each has probability 1/6, so the total probability is 4/6 = 2/3. (b) We can also see that P(D) = 1/6 + 1/6 = 1/3. Since D and Dc are disjoint, P(D) + P(Dc) = 1.
About 9% of people are left-handed. Suppose 2 people are selected at random from the U.S. pop- ulation. Because the sample size of 2 is very small relative to the population, it is reasonable to assume these two people are independent. (a) What is the probability that both are left-handed? (b) What is the probability that both are right-handed?
(a) The probability the first person is left-handed is 0.09, which is the same for the second person. We apply the Multiplication Rule for independent processes to determine the probability that both will be left-handed: 0.09×0.09 = 0.0081. (b) It is reasonable to assume the proportion of people who are ambidextrous (both right- and left-handed) is nearly 0, which results in P (right-handed) = 1 − 0.09 = 0.91. Using the same reasoning as in part (a), the probability that both will be right-handed is 0.91 × 0.91 = 0.8281.
We are interested in the probability of rolling a 1, 4, or 5. (a) Explain why the outcomes 1, 4, and 5 are disjoint. (b) Apply the Addition Rule for disjoint outcomes to determine P (1 or 4 or 5)
(a) The random process is a die roll, and at most one of these outcomes can come up. This means they are disjoint outcomes. (b) P (1 or 4 or 5) = P (1) + P (4) + P (5) = 1 + 1 + 1 = 3 = 1
(a) Determine the probability a randomly drawn loan from the loans data set is from a joint application where the couple had a mortgage. (b) What is the probability that the loan had either of these attributes?
(a) The solution is represented by the intersection of the two circles: 0.095. (b) This is the sum of the three disjoint probabilities shown in the circles: 0.384 + 0.095 + 0.055 = 0.534 (off by 0.001 due to a rounding error).
(a) What is the probability that a randomly selected card is a diamond? (b) What is the probability that a randomly selected card is a face card?
(a) There are 52 cards and 13 diamonds. If the cards are thoroughly shuffled, each card has an equal chance of being drawn, so the probability that a randomly selected card is a diamond is P(♦) = 13 = 0.250. (b) Likewise, there are 12 face cards, so P (face card) = 12 = 3 = 0.231.
About 9% of people are left-handed. Three people are selected at random. (a) What is the probability that the first person is male and right-handed? (b) What is the probability that the first two people are male and right-handed?. (c) What is the probability that the third person is female and left-handed? (d) What is the probability that the first two people are male and right-handed and the third person is female and left-handed?
(a) This can be written in probability notation as P(a randomly selected person is male and right-handed) = 0.455. (b) P(1st 2 are M,RH) = 0.455*0.455 = 0.207. (c) P(3rd F,LH) = 0.5*0.09 = 0.0455. (d) P(1st 2 are M,RH and 3rd F,LH) = 0.0093.
(a) Determine the probability that the algorithm is incorrect if it is known the photo is about fashion. (b) Using the answers from part (a) and 3.29(b), compute P (mach learn is pred fashion | truth is fashion)+ P (mach learn is pred not | truth is fashion) (c) Provide an intuitive argument to explain why the sum in (b) is 1
(a) This probability is P(mach learn is pred not, truth is fashion)/P(truth is fashion) = 0.0615/0.1696 = 0.3626. (b) The total equals 1. (c) Under the condition the photo is about fashion, the ML algorithm must have either predicted it was about fashion or predicted it was not about fashion. The complement still works for conditional probabilities, provided the probabilities are conditioned on the same information.
Cat weights: Question The histogram shown below represents the weights (in kg) of 47 female and 97 male cats
(a) What fraction of these cats weigh less than 2.5 kg? (b) What fraction of these cats weigh between 2.5 and 2.75 kg? (c) What fraction of these cats weigh between 2.75 and 3.5 kg?
In the loans data set in Chapter 2, the homeownership variable described whether the borrower rents, has a mortgage, or owns her property. Of the 10,000 borrowers, 3858 rented, 4789 had a mortgage, and 1353 owned their home.3 (a) Are the outcomes rent, mortgage, and own disjoint? (b) Determine the proportion of loans with value mortgage and own separately. (c) Use the Addition Rule for disjoint outcomes to compute the probability a randomly selected loan from the data set is for someone who has a mortgage or owns her home.
(a) Yes. Each loan is categorized in only one level of homeownership. (b) Mortgage: 4789 = 0.479. Own: 1353 = 0.135. (c) P (mortgage or own) = P (mortgage) + P (own) = 0.479 + 0.135 = 0.614.
(a) Verify the probability of event A, P(A), is 1/3 using the Addition Rule. (b) Do the same for event B
(a)P(A)=P(1or2)=P(1)+P(2)=1+1 =2 =1. (b)Similarly,P(B)=1/3.
After an introductory statistics course, 78% of students can successfully construct tree diagrams. Of those who can construct tree diagrams, 97% passed, while only 57% of those students who could not construct tree diagrams passed. (a) Organize this information into a tree diagram. (b) What is the probability that a randomly selected student passed? (c) Compute the probability a student is able to construct a tree diagram if it is known that she passed.
(b) P (passed) = 0.7566+0.1254 = 0.8820. (c) P (construct tree diagram | passed) = 0.7566/0.8820 = 0.8578.
Probability always takes values between ___ and ___
0 and 1 Probability always takes values between 0 and 1
Use P(inoculated = yes) = 0.0392 and P(result = lived | inoculated = yes) = 0.9754 to determine the probability that a person was both inoculated and lived
0.0382
Jose visits campus every Thursday evening. However, some days the parking garage is full, often due to college events. There are academic events on 35% of evenings, sporting events on 20% of evenings, and no events on 45% of evenings. When there is an academic event, the garage fills up about 25% of the time, and it fills up 70% of evenings with sporting events. On evenings when there are no events, it only fills up about 5% of the time. Q; If Jose comes to campus and finds the garage full, what is the probability that there is a sporting event? -Use *Bayes Theorem* to solve this problem
0.56
What is the chance of getting a 1 or 2 in the next roll?
1 and 2 constitute two of the six equally likely possible outcomes, so the chance of getting one of these two outcomes must be 2/6 = 1/3.
American roulette. The game of American roulette involves spinning a wheel with 38 slots: 18 red, 18 black, and 2 green. A ball is spun onto the wheel and will eventually land in a slot, where each slot has an equal chance of capturing the ball. Gamblers can place bets on red or black. If the ball lands on their color, they double their money. If it lands on another color, they lose their money. Suppose you bet $1 on red. What's the expected value and standard deviation of your winnings?
E = -0.0526. SD = 0.9986.
T/F: Drawing a face card (jack, queen, or king) and drawing a red card from a full deck of playing cards are mutually exclusive events.
False. There are red face cards.
T/F: If a fair coin is tossed many times and the last eight tosses are all heads, then the chance that the next toss will be heads is somewhat less than 50%.
False. These are independent trials.
Ron is watching a roulette table in a casino and notices that the last five outcomes were black. He figures that the chances of getting black six times in a row is very small (about 1/64) and puts his paycheck on red. What is wrong with his reasoning?
He has forgotten that the next roulette spin is independent of the previous spins. Casinos do employ this practice, posting the last several outcomes of many betting games to trick unsuspecting gamblers into believing the odds are in their favor. This is called the gambler's fallacy.
Random processes include rolling a die and flipping a coin. (a) Think of another random process. (b) Describe all the possible outcomes of that process. For instance, rolling a die is a random process with possible outcomes 1, 2, ..., 6.
Here are four examples. (i) Whether someone gets sick in the next month or not is an apparently random process with outcomes sick and not. (ii) We can generate a random process by randomly picking a person and measuring that person's height. The outcome of this process will be a positive number. (iii) Whether the stock market goes up or down next week is a seemingly random process with possible outcomes up, down, and no change. Alternatively, we could have used the percent change in the stock market as a numerical outcome. (iv) Whether your roommate cleans her dishes tonight probably seems like a random process with possible outcomes cleans dishes and leaves dishes.
Consider rolling two dice. If 1/6 of the time the first die is a 1 and 1/6 of those times the second die is a 1, what is the chance of getting two 1s?
If 16. ̄6% of the time the first die is a 1 and 1/6 of those times the second die is also a 1, then the chance that both dice are 1 is (1/6) × (1/6) or 1/36.
We sample a photo from the data set and learn the ML algorithm predicted this photo was not about fashion. What is the probability that it was incorrect and the photo is about fashion?
If the ML classifier suggests a photo is not about fashion, then it comes from the second row in the data set. Of these 1603 photos, 112 were actually about fashion: P (truth is fashion given mach learn is pred not) = 112/1603 = 0.070
A "die", the singular of dice, is a cube with six faces numbered 1, 2, 3, 4, 5, and 6. What is the chance of getting 1 when rolling a die?
If the die is fair, then the chance of a 1 is as good as the chance of any other number. Since there are six outcomes, the chance must be 1-in-6 or, equivalently, 1/6.
Consider the midterm and final for a statistics class. Suppose 13% of students earned an A on the midterm. Of those students who earned an A on the midterm, 47% received an A on the final, and 11% of the students who earned lower than an A on the midterm received an A on the final. You randomly pick up a final exam and notice the student received an A. What is the probability that this student earned an A on the midterm?
P (midterm = A | final = A) = P (midterm = A and final = A)/P(final = A) = 0.0611/(0.0611+0.0957) =0.3897
In Guided Practice 3.43 and 3.45, you found that if the parking lot is full, the probability there is a sporting event is 0.56 and the probability there is an academic event is 0.35. Using this information, compute P (no event | the lot is full).
P (no event | the lot is full) = 1 - P (academic event | the lot is full) - P (sporting event | the lot is full) = 1-0.35-0.56 = 0.09 = 9%
Consider the smallpox data set. Suppose we are given only two pieces of information: 96.08% of residents were not inoculated, and 85.88% of the residents who were not inoculated ended up surviving. How could we compute the probability that a resident was not inoculated and lived?
P (result = lived and inoculated = no) and we are given thatP (result = lived | inoculated = no) = 0.8588 P (inoculated = no) = 0.9608 Among the 96.08% of people who were not inoculated, 85.88% survived:P (result = lived and inoculated = no) = 0.8588 × 0.9608 = 0.8251 This is equivalent to the General Multiplication Rule.
Student outfits. In a classroom with 24 students, 7 students are wearing jeans, 4 are wearing shorts, 8 are wearing skirts, and the rest are wearing leggings. If we randomly select 3 students without replacement, what is the probability that one of the selected students is wearing leggings and the other two are wearing jeans? Note that these are mutually exclusive clothing options.
P(1leggings, 2jeans, 3jeans) = 5/24 × 7/23 × 6/22 = 0.0173. However, the person with leggings could have come 2nd or 3rd, and these each have this same probability, so 3 × 0.0173 = 0.0519.
Suppose 80% of people like peanut butter, 89% like jelly, and 78% like both. Given that a randomly sampled person likes peanut butter, what's the probability that he also likes jelly?
P(J+ | PB+) = P(PB+ and J+)/P(J+) = 0.78/0.80 = 0.975
It's never lupus. Lupus is a medical phenomenon where antibodies that are supposed to attack foreign cells to prevent infections instead see plasma proteins as foreign bodies, leading to a high risk of blood clotting. It is believed that 2% of the population suffer from this disease. The test is 98% accurate if a person actually has the disease. The test is 74% accurate if a person does not have the disease. There is a line from the Fox television show House that is often used after a patient tests positive for lupus: "It's never lupus." Do you think there is truth to this statement? Use appropriate probabilities to support your answer.
P(Lupus | + test) = (0.02*0.98)/[(0.02*0.98)+(0.98*0.26)] = 0.0714 Lupus 7% of the time with a positive test.
Exit poll. Edison Research gathered exit poll results from several sources for the Wisconsin recall election of Scott Walker. They found that 53% of the respondents voted in favor of Scott Walker. Additionally, they estimated that of those who did vote in favor for Scott Walker, 37% had a college degree, while 44% of those who voted against Scott Walker had a college degree. Suppose we randomly sampled a person who participated in the exit poll and found that he had a college degree. What is the probability that he voted in favor of Scott Walker?
P(Walker | Degree) = (0.53*0.37)/[(0.53*0.37)+(0.47*0.44)] =0.48672 = 49% change they voted for walker
Jose visits campus every Thursday evening. However, some days the parking garage is full, often due to college events. There are academic events on 35% of evenings, sporting events on 20% of evenings, and no events on 45% of evenings. When there is an academic event, the garage fills up about 25% of the time, and it fills up 70% of evenings with sporting events. On evenings when there are no events, it only fills up about 5% of the time. Q: What is the probability that there is an academic event conditioned on the parking lot being full?
P(academic | full) = P(academic and full)/P(full) =0.0875/(0.0875+0.014+0.025) = 0.35
In Canada, about 0.35% of women over 40 will develop breast cancer in any given year. A common screening test for cancer is the mammogram, but this test is not perfect. In about 11% of patients with breast cancer, the test gives a false negative: it indicates a woman does not have breast cancer when she does have breast cancer. Similarly, the test gives a false positive in 7% of patients who do not have breast cancer: it indicates these patients have breast cancer when they actually do not. If we tested a random woman over 40 for breast cancer using a mammogram and the test came back positive - that is, the test suggested the patient has cancer - what is the probability that the patient actually has breast cancer?
P(has BC | mammogram+) = P(has BC and mammogram+)/P (mammogram+ ) = 0.00312/0.07288 ~0.0428
Backgammon. Backgammon is a board game for two players in which the playing pieces are moved according to the roll of two dice. Players win by removing all of their pieces from the board, so it is usually good to roll high numbers. You are playing backgammon with a friend and you roll two 6s in your first roll and two 6s in your second roll. Your friend rolls two 3s in his first roll and again in his second row. Your friend claims that you are cheating, because rolling double 6s twice in a row is very unlikely. Using probability, show that your rolls were just as likely as his.
P(roll a 6) = 1/6 P( roll a 6 and roll a 6) = 1/6*1/6 = 1/36 P(roll a 3) = 1/6 P( roll a 3 and roll a 3) = 1/6*1/6 = 1/36
Jose visits campus every Thursday evening. However, some days the parking garage is full, often due to college events. There are academic events on 35% of evenings, sporting events on 20% of evenings, and no events on 45% of evenings. When there is an academic event, the garage fills up about 25% of the time, and it fills up 70% of evenings with sporting events. On evenings when there are no events, it only fills up about 5% of the time. Q: If Jose comes to campus and finds the garage full, what is the probability that there is a sporting event? -Use a *tree diagram* to solve this problem
P(sporting | full) = P(sporting and full)/P(full) =0.14/(0.0875+0.014+0.025) = 0.56
Compute the probability that event B or event D occurs.
Since B and D are disjoint events, use the Addition Rule: P(BorD)=P(B)+P(D)=1+1 =2.333
What is the chance of not rolling a 2?
Since the chance of rolling a 2 is 1/6 or 16. ̄6%, the chance of not rolling a 2 must be 100%−16. ̄6% = 83. ̄3% or 5/6. Alternatively, we could have noticed that not rolling a 2 is the same as getting a 1, 3, 4, 5, or 6, which makes up five of the six equally likely outcomes and has probability 5/6.
Figure 3.6 suggests three distributions for household income in the United States. Only one is correct. Which one must it be? What is wrong with the other two?
The probabilities of (a) do not sum to 1. The second probability in (b) is negative. This leaves (c), which sure enough satisfies the requirements of a distribution. One of the three was said to be the actual distribution of US household incomes, so it must be (c)
If we shuffle up a deck of cards and draw one, is the event that the card is a heart independent of the event that the card is an ace?
The probability the card is a heart is 1/4 and the probability that it is an ace is 1/13. The probability the card is the ace of hearts is 1/52. We check whether P (A and B) = P (A) × P (B) is satisfied: P(♥)×P(ace)=1/4 × 1/31 = 1/52 =P(♥ and ace) Because the equation holds, the event that the card is a heart and the event that the card is an ace are independent events.
What is the probability of rolling the three dice and getting all 1s?
The same logic applies from the 2 dice example. P (die1 = 1 and die2 = 1 and die3 = 1) = P (die1 = 1) × P (die2 = 1) × P (die3 = 1) = (1/6) × (1/6) × (1/6) = 1/216
Based on the probabilities computed above, does it appear that inoculation is effective at reducing the risk of death from smallpox?
The samples are large relative to the difference in death rates for the "inoculated" and "not inoculated" groups, so it seems there is an association between inoculated and outcome. However, this is an observational study and we cannot be sure if there is a causal connection. (Further research has shown that inoculation is effective at reducing death rates.)
If 97.54% of the inoculated people lived, what proportion of inoculated people must have died?
There were only two possible outcomes: lived or died. This means that 100% - 97.54% = 2.46% of the people who were inoculated died.
Suppose a person's height is rounded to the nearest centimeter. Is there a chance that a random person's measured height will be 180 cm?
This has positive probability. Anyone between 179.5 cm and 180.5 cm will have a measured height of 180 cm
What is the probability that a randomly selected person is exactly 180 cm? Assume you can measure perfectly.
This probability is zero. A person might be close to 180 cm, but not exactly 180 cm tall. This also makes sense with the definition of probability as area; there is no area captured between 180 cm and 180 cm.
T/F: Drawing a face card and drawing an ace from a full deck of playing cards are mutually exclusive events.
True. A card cannot be both a face card and an ace.
If a photo is actually about fashion, what is the chance the ML classifier correctly identified the photo as being about fashion?
We can estimate this probability using the data. Of the 309 fashion photos, the ML algorithm correctly classified 197 of the photos: P (mach learn is pred fashion given truth is fashion) = 197/309 = 0.638
It takes John an average of 18 minutes each day to commute to work. What would you expect his average commute time to be for the week? Would you be surprised if John's weekly commute wasn't exactly 90 minutes
We were told that the average (i.e. expected value) of the commute time is 18 minutes per day: E(Xi) = 18. To get the expected time for the sum of the five days, we can add up the expected time for each individual day: E(W)=E(X1 +X2 +X3 +X4 +X5)= E(X1) + E(X2) + E(X3) + E(X4) + E(X5) = 18 + 18 + 18 + 18 + 18 = 90 minutes No, since there is probably some variability. For example, the traffic will vary from one day to next
Suppose the professor randomly picks from a class of 15 students without regard to who she already selected, i.e. students can be picked more than once. a) What is the probability that you will not be picked for any of the three questions? b) what is the probability of being picked to answer all three ques- tions?
a) Each pick is independent, and the probability of not being picked for any individual question is 14/15. Thus, we can use the Multiplication Rule for independent processes. You have a slightly higher chance of not being picked compared to when she picked a new person for each question. However, you now may be picked more than once. b) P(being picked to answer all three questions) = (1/15)^3 = 0.00030.
Answer: Global Warming
a) No, 0.18 of respondents fall into this combination. (b) 0.60 + 0.20 − 0.18 = 0.62. (c) 0.18/0.20 = 0.9. (d) 0.11/0.33 ≈ 0.33. (e) No, otherwise the answers to (c) and (d) would be the same. (f) 0.06/0.34 ≈ 0.18.
Answer: Health coverage
a) No, 0.2099 of respondents fall into this combination. b) P(Excellent Health) = 0.2329 c) =0.2099/0.8738 = 0.2402 d) =0.0230/0.1262 =0.18225 e) No, otherwise (c) and (d) would be the same
